The angular speed of the ball just after the blow.

Question

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Answer 1

Question

Chapter 9, Problem 106P

(a)

To determine

The angular speed of the ball just after the blow.

(a)

Expert Solution

Answer to Problem 106P

$ω0=−5v06R$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline

Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

Newton’s second law of motion in rotational form

$∑τ=r×F=Icmα$

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 1

FIGURE: 1

Applying Newton’s second law in rotational form to ball,

$∑τ0=Fav(h−R)=Icmα=Icmω0Δt$

$⇒Fav(h−R)=Icmω0Δt$

$ω0=Fav(h−R)ΔtIcm$ $(1)$

Where,

$τ0$ is the torque about the center of the ball

$Fav$ is the average force on the ball

$h$ is the below the centerline at which the force is applied on the ball

$R$ is radius of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$ω0$ is angular speed of the ball after impact

$Δt$ is elapsed time

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mr2$

Substituting this in equation $(1)$ ,

$ω0=Fav(h−r)Δt25mr2$ $(2)$

Applying impulse-momentum theorem to the ball,

$I=FavΔt=Δp=mv0$

$⇒mv0=FavΔt$

$v0=FavΔtm$ $(3)$

Where, $m$ is mass of the ball

$Δp$ is momentum, and

$v0$ is speed of the ball just after the impact

From equation $(3)$ , $v0=FavΔtm⇒Δt=mv0Fav$

Substituting the expression for $Δt$ in equation $(3)$ ,

$ω0=Fav(h−R)mv0Fav25mR2=5v0(h−R)2R2$

Substituting $h=2R3$ in the above equation,

$ω0=5v0(2R3−R)2R2=5v0(2R−3R3)2R2=5v0(−R3)2R2=−5v06R$

Conclusion:

The angular speed of the ball just after the blow is $−5v06R$ .

(b)

To determine

The speed of the ball once it begins to roll without slipping.

(b)

Expert Solution

Answer to Problem 106P

$v=521v0$

Explanation of Solution

Given: Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 2

FIGURE: 2

Referring to the force diagram shown in figure 2, applying Newton’s second law to the ball when it is rolling without slipping,

$Στ0=fkR=Icmα$ $(4)$

$ΣFy=Fn−mg=0$ $(5)$

And

$ΣFx=−fk=ma$ $(6)$

Where,

$Fx$ is force in the x-direction

$fk$ is force of friction in x-direction

$Fy$ is force in the y-direction

$fn$ is force of friction in y-direction

$m$ is mass of the ball

$a$ is the acceleration of the ball

$g$ is acceleration due to gravity

$R$ is radius of the ball

$τ0$ is torque about the center of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$g$ is acceleration due to gravity, $g=9.8 m/s2$

But, $fk=μkFn$ $(7)$

Where, $μk$ is coefficient of kinetic friction

Calculations:

From equation $(5)$ , $Fn=mg$

Substituting this in equation $(7)$ ,

$fk=μkmg$

From equation $(4)$ angular acceleration,

$α=fkRIcm$

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mR2$

Substituting for $fk$ and $Icm$ ,

$α=μkmgR25mR2=5μkg2R$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

$ω=ω0+αΔt=ω0+5μkg2RΔt$ $(8)$

Now,substituting the expression for $fk$ in equation $(6)$ ,

$−μkmg=ma⇒a=−μkg$ $(9)$

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

$v=v0+aΔt$ $(10)$

Where,

$v$ is speed of the ball

$v0$ is speed of the ball just after impact

$a$ is the acceleration ball

$Δt$ is elapsed time

Substituting for $a$ from equation $(9)$ in equation $(10)$ ,

$v=v0−μkgΔt$ $(11)$

Imposing the condition for rolling the ball without slipping,

$R(ω0+5μkg2RΔt)=v0−μkgΔt⇒Δt=1621v0μkg$

Substituting this $Δt$ in equation $(11)$ ,

$v=v0−μkg(1621v0μkg)=521v0$

Conclusion:

The speed of the ball once it begins to roll without slipping is $521v0$ .

(c)

To determine

The kinetic energy of the ball just after the hit.

(c)

Expert Solution

Answer to Problem 106P

$Ki=23mv0236$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline.

Speed of the ball just after the blow is $v0$ .

The coefficient of kinetic friction between the ball and billiard table is $μk$ .

Formula Used:

Initial kinetic energy of the ball can be written as,

$Ki=Ktrans+Krot$ $(12)$

Where, $Ktrans$ is kinetic energy due to translational motion of the ball, $Ktrans=12mv02$

$Krot$ is kinetic energy due to rotational energy of the ball, $Krot=12Iω02$

Substituting the expressions for $Ktrans$ and $Krot$ in equation $(12)$ ,

$Ki=12mv02+12Iω02$ $(13)$

Where, $m$ is mass of the ball,

$v0$ is speed of the ball just after impact

$I$ is moment of inertia of the ball

$ω0$ is angular speed of the ball

Moment of inertia, $I=25mR2$ , and Angular speed, $ω0=−5v06R$

Substituting these in equation $(13)$ ,

$Ki=12mv02+12(25mR2)(−5v06R)2$

$=12mv02+15mR2×2536v02R2$

$=12mv02+536mv02$

$=18mv02+5mv0236=23mv0236$

Conclusion:

The kinetic energy of the ball just after the hit is $23mv0236$ .

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Answer 2

Question

Chapter 9, Problem 106P

(a)

To determine

The angular speed of the ball just after the blow.

(a)

Expert Solution

Answer to Problem 106P

$ω0=−5v06R$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline

Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

Newton’s second law of motion in rotational form

$∑τ=r×F=Icmα$

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 1

FIGURE: 1

Applying Newton’s second law in rotational form to ball,

$∑τ0=Fav(h−R)=Icmα=Icmω0Δt$

$⇒Fav(h−R)=Icmω0Δt$

$ω0=Fav(h−R)ΔtIcm$ $(1)$

Where,

$τ0$ is the torque about the center of the ball

$Fav$ is the average force on the ball

$h$ is the below the centerline at which the force is applied on the ball

$R$ is radius of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$ω0$ is angular speed of the ball after impact

$Δt$ is elapsed time

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mr2$

Substituting this in equation $(1)$ ,

$ω0=Fav(h−r)Δt25mr2$ $(2)$

Applying impulse-momentum theorem to the ball,

$I=FavΔt=Δp=mv0$

$⇒mv0=FavΔt$

$v0=FavΔtm$ $(3)$

Where, $m$ is mass of the ball

$Δp$ is momentum, and

$v0$ is speed of the ball just after the impact

From equation $(3)$ , $v0=FavΔtm⇒Δt=mv0Fav$

Substituting the expression for $Δt$ in equation $(3)$ ,

$ω0=Fav(h−R)mv0Fav25mR2=5v0(h−R)2R2$

Substituting $h=2R3$ in the above equation,

$ω0=5v0(2R3−R)2R2=5v0(2R−3R3)2R2=5v0(−R3)2R2=−5v06R$

Conclusion:

The angular speed of the ball just after the blow is $−5v06R$ .

(b)

To determine

The speed of the ball once it begins to roll without slipping.

(b)

Expert Solution

Answer to Problem 106P

$v=521v0$

Explanation of Solution

Given: Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 2

FIGURE: 2

Referring to the force diagram shown in figure 2, applying Newton’s second law to the ball when it is rolling without slipping,

$Στ0=fkR=Icmα$ $(4)$

$ΣFy=Fn−mg=0$ $(5)$

And

$ΣFx=−fk=ma$ $(6)$

Where,

$Fx$ is force in the x-direction

$fk$ is force of friction in x-direction

$Fy$ is force in the y-direction

$fn$ is force of friction in y-direction

$m$ is mass of the ball

$a$ is the acceleration of the ball

$g$ is acceleration due to gravity

$R$ is radius of the ball

$τ0$ is torque about the center of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$g$ is acceleration due to gravity, $g=9.8 m/s2$

But, $fk=μkFn$ $(7)$

Where, $μk$ is coefficient of kinetic friction

Calculations:

From equation $(5)$ , $Fn=mg$

Substituting this in equation $(7)$ ,

$fk=μkmg$

From equation $(4)$ angular acceleration,

$α=fkRIcm$

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mR2$

Substituting for $fk$ and $Icm$ ,

$α=μkmgR25mR2=5μkg2R$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

$ω=ω0+αΔt=ω0+5μkg2RΔt$ $(8)$

Now,substituting the expression for $fk$ in equation $(6)$ ,

$−μkmg=ma⇒a=−μkg$ $(9)$

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

$v=v0+aΔt$ $(10)$

Where,

$v$ is speed of the ball

$v0$ is speed of the ball just after impact

$a$ is the acceleration ball

$Δt$ is elapsed time

Substituting for $a$ from equation $(9)$ in equation $(10)$ ,

$v=v0−μkgΔt$ $(11)$

Imposing the condition for rolling the ball without slipping,

$R(ω0+5μkg2RΔt)=v0−μkgΔt⇒Δt=1621v0μkg$

Substituting this $Δt$ in equation $(11)$ ,

$v=v0−μkg(1621v0μkg)=521v0$

Conclusion:

The speed of the ball once it begins to roll without slipping is $521v0$ .

(c)

To determine

The kinetic energy of the ball just after the hit.

(c)

Expert Solution

Answer to Problem 106P

$Ki=23mv0236$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline.

Speed of the ball just after the blow is $v0$ .

The coefficient of kinetic friction between the ball and billiard table is $μk$ .

Formula Used:

Initial kinetic energy of the ball can be written as,

$Ki=Ktrans+Krot$ $(12)$

Where, $Ktrans$ is kinetic energy due to translational motion of the ball, $Ktrans=12mv02$

$Krot$ is kinetic energy due to rotational energy of the ball, $Krot=12Iω02$

Substituting the expressions for $Ktrans$ and $Krot$ in equation $(12)$ ,

$Ki=12mv02+12Iω02$ $(13)$

Where, $m$ is mass of the ball,

$v0$ is speed of the ball just after impact

$I$ is moment of inertia of the ball

$ω0$ is angular speed of the ball

Moment of inertia, $I=25mR2$ , and Angular speed, $ω0=−5v06R$

Substituting these in equation $(13)$ ,

$Ki=12mv02+12(25mR2)(−5v06R)2$

$=12mv02+15mR2×2536v02R2$

$=12mv02+536mv02$

$=18mv02+5mv0236=23mv0236$

Conclusion:

The kinetic energy of the ball just after the hit is $23mv0236$ .

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Answer 3

Question

Chapter 9, Problem 106P

(a)

To determine

The angular speed of the ball just after the blow.

(a)

Expert Solution

Answer to Problem 106P

$ω0=−5v06R$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline

Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

Newton’s second law of motion in rotational form

$∑τ=r×F=Icmα$

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 1

FIGURE: 1

Applying Newton’s second law in rotational form to ball,

$∑τ0=Fav(h−R)=Icmα=Icmω0Δt$

$⇒Fav(h−R)=Icmω0Δt$

$ω0=Fav(h−R)ΔtIcm$ $(1)$

Where,

$τ0$ is the torque about the center of the ball

$Fav$ is the average force on the ball

$h$ is the below the centerline at which the force is applied on the ball

$R$ is radius of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$ω0$ is angular speed of the ball after impact

$Δt$ is elapsed time

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mr2$

Substituting this in equation $(1)$ ,

$ω0=Fav(h−r)Δt25mr2$ $(2)$

Applying impulse-momentum theorem to the ball,

$I=FavΔt=Δp=mv0$

$⇒mv0=FavΔt$

$v0=FavΔtm$ $(3)$

Where, $m$ is mass of the ball

$Δp$ is momentum, and

$v0$ is speed of the ball just after the impact

From equation $(3)$ , $v0=FavΔtm⇒Δt=mv0Fav$

Substituting the expression for $Δt$ in equation $(3)$ ,

$ω0=Fav(h−R)mv0Fav25mR2=5v0(h−R)2R2$

Substituting $h=2R3$ in the above equation,

$ω0=5v0(2R3−R)2R2=5v0(2R−3R3)2R2=5v0(−R3)2R2=−5v06R$

Conclusion:

The angular speed of the ball just after the blow is $−5v06R$ .

(b)

To determine

The speed of the ball once it begins to roll without slipping.

(b)

Expert Solution

Answer to Problem 106P

$v=521v0$

Explanation of Solution

Given: Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 2

FIGURE: 2

Referring to the force diagram shown in figure 2, applying Newton’s second law to the ball when it is rolling without slipping,

$Στ0=fkR=Icmα$ $(4)$

$ΣFy=Fn−mg=0$ $(5)$

And

$ΣFx=−fk=ma$ $(6)$

Where,

$Fx$ is force in the x-direction

$fk$ is force of friction in x-direction

$Fy$ is force in the y-direction

$fn$ is force of friction in y-direction

$m$ is mass of the ball

$a$ is the acceleration of the ball

$g$ is acceleration due to gravity

$R$ is radius of the ball

$τ0$ is torque about the center of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$g$ is acceleration due to gravity, $g=9.8 m/s2$

But, $fk=μkFn$ $(7)$

Where, $μk$ is coefficient of kinetic friction

Calculations:

From equation $(5)$ , $Fn=mg$

Substituting this in equation $(7)$ ,

$fk=μkmg$

From equation $(4)$ angular acceleration,

$α=fkRIcm$

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mR2$

Substituting for $fk$ and $Icm$ ,

$α=μkmgR25mR2=5μkg2R$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

$ω=ω0+αΔt=ω0+5μkg2RΔt$ $(8)$

Now,substituting the expression for $fk$ in equation $(6)$ ,

$−μkmg=ma⇒a=−μkg$ $(9)$

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

$v=v0+aΔt$ $(10)$

Where,

$v$ is speed of the ball

$v0$ is speed of the ball just after impact

$a$ is the acceleration ball

$Δt$ is elapsed time

Substituting for $a$ from equation $(9)$ in equation $(10)$ ,

$v=v0−μkgΔt$ $(11)$

Imposing the condition for rolling the ball without slipping,

$R(ω0+5μkg2RΔt)=v0−μkgΔt⇒Δt=1621v0μkg$

Substituting this $Δt$ in equation $(11)$ ,

$v=v0−μkg(1621v0μkg)=521v0$

Conclusion:

The speed of the ball once it begins to roll without slipping is $521v0$ .

(c)

To determine

The kinetic energy of the ball just after the hit.

(c)

Expert Solution

Answer to Problem 106P

$Ki=23mv0236$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline.

Speed of the ball just after the blow is $v0$ .

The coefficient of kinetic friction between the ball and billiard table is $μk$ .

Formula Used:

Initial kinetic energy of the ball can be written as,

$Ki=Ktrans+Krot$ $(12)$

Where, $Ktrans$ is kinetic energy due to translational motion of the ball, $Ktrans=12mv02$

$Krot$ is kinetic energy due to rotational energy of the ball, $Krot=12Iω02$

Substituting the expressions for $Ktrans$ and $Krot$ in equation $(12)$ ,

$Ki=12mv02+12Iω02$ $(13)$

Where, $m$ is mass of the ball,

$v0$ is speed of the ball just after impact

$I$ is moment of inertia of the ball

$ω0$ is angular speed of the ball

Moment of inertia, $I=25mR2$ , and Angular speed, $ω0=−5v06R$

Substituting these in equation $(13)$ ,

$Ki=12mv02+12(25mR2)(−5v06R)2$

$=12mv02+15mR2×2536v02R2$

$=12mv02+536mv02$

$=18mv02+5mv0236=23mv0236$

Conclusion:

The kinetic energy of the ball just after the hit is $23mv0236$ .

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Answer 4

Question

Chapter 9, Problem 106P

(a)

To determine

The angular speed of the ball just after the blow.

(a)

Expert Solution

Answer to Problem 106P

$ω0=−5v06R$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline

Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

Newton’s second law of motion in rotational form

$∑τ=r×F=Icmα$

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 1

FIGURE: 1

Applying Newton’s second law in rotational form to ball,

$∑τ0=Fav(h−R)=Icmα=Icmω0Δt$

$⇒Fav(h−R)=Icmω0Δt$

$ω0=Fav(h−R)ΔtIcm$ $(1)$

Where,

$τ0$ is the torque about the center of the ball

$Fav$ is the average force on the ball

$h$ is the below the centerline at which the force is applied on the ball

$R$ is radius of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$ω0$ is angular speed of the ball after impact

$Δt$ is elapsed time

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mr2$

Substituting this in equation $(1)$ ,

$ω0=Fav(h−r)Δt25mr2$ $(2)$

Applying impulse-momentum theorem to the ball,

$I=FavΔt=Δp=mv0$

$⇒mv0=FavΔt$

$v0=FavΔtm$ $(3)$

Where, $m$ is mass of the ball

$Δp$ is momentum, and

$v0$ is speed of the ball just after the impact

From equation $(3)$ , $v0=FavΔtm⇒Δt=mv0Fav$

Substituting the expression for $Δt$ in equation $(3)$ ,

$ω0=Fav(h−R)mv0Fav25mR2=5v0(h−R)2R2$

Substituting $h=2R3$ in the above equation,

$ω0=5v0(2R3−R)2R2=5v0(2R−3R3)2R2=5v0(−R3)2R2=−5v06R$

Conclusion:

The angular speed of the ball just after the blow is $−5v06R$ .

(b)

To determine

The speed of the ball once it begins to roll without slipping.

(b)

Expert Solution

Answer to Problem 106P

$v=521v0$

Explanation of Solution

Given: Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 2

FIGURE: 2

Referring to the force diagram shown in figure 2, applying Newton’s second law to the ball when it is rolling without slipping,

$Στ0=fkR=Icmα$ $(4)$

$ΣFy=Fn−mg=0$ $(5)$

And

$ΣFx=−fk=ma$ $(6)$

Where,

$Fx$ is force in the x-direction

$fk$ is force of friction in x-direction

$Fy$ is force in the y-direction

$fn$ is force of friction in y-direction

$m$ is mass of the ball

$a$ is the acceleration of the ball

$g$ is acceleration due to gravity

$R$ is radius of the ball

$τ0$ is torque about the center of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$g$ is acceleration due to gravity, $g=9.8 m/s2$

But, $fk=μkFn$ $(7)$

Where, $μk$ is coefficient of kinetic friction

Calculations:

From equation $(5)$ , $Fn=mg$

Substituting this in equation $(7)$ ,

$fk=μkmg$

From equation $(4)$ angular acceleration,

$α=fkRIcm$

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mR2$

Substituting for $fk$ and $Icm$ ,

$α=μkmgR25mR2=5μkg2R$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

$ω=ω0+αΔt=ω0+5μkg2RΔt$ $(8)$

Now,substituting the expression for $fk$ in equation $(6)$ ,

$−μkmg=ma⇒a=−μkg$ $(9)$

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

$v=v0+aΔt$ $(10)$

Where,

$v$ is speed of the ball

$v0$ is speed of the ball just after impact

$a$ is the acceleration ball

$Δt$ is elapsed time

Substituting for $a$ from equation $(9)$ in equation $(10)$ ,

$v=v0−μkgΔt$ $(11)$

Imposing the condition for rolling the ball without slipping,

$R(ω0+5μkg2RΔt)=v0−μkgΔt⇒Δt=1621v0μkg$

Substituting this $Δt$ in equation $(11)$ ,

$v=v0−μkg(1621v0μkg)=521v0$

Conclusion:

The speed of the ball once it begins to roll without slipping is $521v0$ .

(c)

To determine

The kinetic energy of the ball just after the hit.

(c)

Expert Solution

Answer to Problem 106P

$Ki=23mv0236$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline.

Speed of the ball just after the blow is $v0$ .

The coefficient of kinetic friction between the ball and billiard table is $μk$ .

Formula Used:

Initial kinetic energy of the ball can be written as,

$Ki=Ktrans+Krot$ $(12)$

Where, $Ktrans$ is kinetic energy due to translational motion of the ball, $Ktrans=12mv02$

$Krot$ is kinetic energy due to rotational energy of the ball, $Krot=12Iω02$

Substituting the expressions for $Ktrans$ and $Krot$ in equation $(12)$ ,

$Ki=12mv02+12Iω02$ $(13)$

Where, $m$ is mass of the ball,

$v0$ is speed of the ball just after impact

$I$ is moment of inertia of the ball

$ω0$ is angular speed of the ball

Moment of inertia, $I=25mR2$ , and Angular speed, $ω0=−5v06R$

Substituting these in equation $(13)$ ,

$Ki=12mv02+12(25mR2)(−5v06R)2$

$=12mv02+15mR2×2536v02R2$

$=12mv02+536mv02$

$=18mv02+5mv0236=23mv0236$

Conclusion:

The kinetic energy of the ball just after the hit is $23mv0236$ .

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Answer 5

Chapter 9, Problem 106P

(a)

To determine

The angular speed of the ball just after the blow.

(a)

Expert Solution

Answer to Problem 106P

$ω0=−5v06R$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline

Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

Newton’s second law of motion in rotational form

$∑τ=r×F=Icmα$

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 1

FIGURE: 1

Applying Newton’s second law in rotational form to ball,

$∑τ0=Fav(h−R)=Icmα=Icmω0Δt$

$⇒Fav(h−R)=Icmω0Δt$

$ω0=Fav(h−R)ΔtIcm$ $(1)$

Where,

$τ0$ is the torque about the center of the ball

$Fav$ is the average force on the ball

$h$ is the below the centerline at which the force is applied on the ball

$R$ is radius of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$ω0$ is angular speed of the ball after impact

$Δt$ is elapsed time

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mr2$

Substituting this in equation $(1)$ ,

$ω0=Fav(h−r)Δt25mr2$ $(2)$

Applying impulse-momentum theorem to the ball,

$I=FavΔt=Δp=mv0$

$⇒mv0=FavΔt$

$v0=FavΔtm$ $(3)$

Where, $m$ is mass of the ball

$Δp$ is momentum, and

$v0$ is speed of the ball just after the impact

From equation $(3)$ , $v0=FavΔtm⇒Δt=mv0Fav$

Substituting the expression for $Δt$ in equation $(3)$ ,

$ω0=Fav(h−R)mv0Fav25mR2=5v0(h−R)2R2$

Substituting $h=2R3$ in the above equation,

$ω0=5v0(2R3−R)2R2=5v0(2R−3R3)2R2=5v0(−R3)2R2=−5v06R$

Conclusion:

The angular speed of the ball just after the blow is $−5v06R$ .

(b)

To determine

The speed of the ball once it begins to roll without slipping.

(b)

Expert Solution

Answer to Problem 106P

$v=521v0$

Explanation of Solution

Given: Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 2

FIGURE: 2

Referring to the force diagram shown in figure 2, applying Newton’s second law to the ball when it is rolling without slipping,

$Στ0=fkR=Icmα$ $(4)$

$ΣFy=Fn−mg=0$ $(5)$

And

$ΣFx=−fk=ma$ $(6)$

Where,

$Fx$ is force in the x-direction

$fk$ is force of friction in x-direction

$Fy$ is force in the y-direction

$fn$ is force of friction in y-direction

$m$ is mass of the ball

$a$ is the acceleration of the ball

$g$ is acceleration due to gravity

$R$ is radius of the ball

$τ0$ is torque about the center of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$g$ is acceleration due to gravity, $g=9.8 m/s2$

But, $fk=μkFn$ $(7)$

Where, $μk$ is coefficient of kinetic friction

Calculations:

From equation $(5)$ , $Fn=mg$

Substituting this in equation $(7)$ ,

$fk=μkmg$

From equation $(4)$ angular acceleration,

$α=fkRIcm$

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mR2$

Substituting for $fk$ and $Icm$ ,

$α=μkmgR25mR2=5μkg2R$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

$ω=ω0+αΔt=ω0+5μkg2RΔt$ $(8)$

Now,substituting the expression for $fk$ in equation $(6)$ ,

$−μkmg=ma⇒a=−μkg$ $(9)$

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

$v=v0+aΔt$ $(10)$

Where,

$v$ is speed of the ball

$v0$ is speed of the ball just after impact

$a$ is the acceleration ball

$Δt$ is elapsed time

Substituting for $a$ from equation $(9)$ in equation $(10)$ ,

$v=v0−μkgΔt$ $(11)$

Imposing the condition for rolling the ball without slipping,

$R(ω0+5μkg2RΔt)=v0−μkgΔt⇒Δt=1621v0μkg$

Substituting this $Δt$ in equation $(11)$ ,

$v=v0−μkg(1621v0μkg)=521v0$

Conclusion:

The speed of the ball once it begins to roll without slipping is $521v0$ .

(c)

To determine

The kinetic energy of the ball just after the hit.

(c)

Expert Solution

Answer to Problem 106P

$Ki=23mv0236$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline.

Speed of the ball just after the blow is $v0$ .

The coefficient of kinetic friction between the ball and billiard table is $μk$ .

Formula Used:

Initial kinetic energy of the ball can be written as,

$Ki=Ktrans+Krot$ $(12)$

Where, $Ktrans$ is kinetic energy due to translational motion of the ball, $Ktrans=12mv02$

$Krot$ is kinetic energy due to rotational energy of the ball, $Krot=12Iω02$

Substituting the expressions for $Ktrans$ and $Krot$ in equation $(12)$ ,

$Ki=12mv02+12Iω02$ $(13)$

Where, $m$ is mass of the ball,

$v0$ is speed of the ball just after impact

$I$ is moment of inertia of the ball

$ω0$ is angular speed of the ball

Moment of inertia, $I=25mR2$ , and Angular speed, $ω0=−5v06R$

Substituting these in equation $(13)$ ,

$Ki=12mv02+12(25mR2)(−5v06R)2$

$=12mv02+15mR2×2536v02R2$

$=12mv02+536mv02$

$=18mv02+5mv0236=23mv0236$

Conclusion:

The kinetic energy of the ball just after the hit is $23mv0236$ .

Answer 6

Chapter 9, Problem 106P

(a)

To determine

The angular speed of the ball just after the blow.

(a)

Expert Solution

Answer to Problem 106P

$ω0=−5v06R$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline

Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

Newton’s second law of motion in rotational form

$∑τ=r×F=Icmα$

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 1

FIGURE: 1

Applying Newton’s second law in rotational form to ball,

$∑τ0=Fav(h−R)=Icmα=Icmω0Δt$

$⇒Fav(h−R)=Icmω0Δt$

$ω0=Fav(h−R)ΔtIcm$ $(1)$

Where,

$τ0$ is the torque about the center of the ball

$Fav$ is the average force on the ball

$h$ is the below the centerline at which the force is applied on the ball

$R$ is radius of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$ω0$ is angular speed of the ball after impact

$Δt$ is elapsed time

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mr2$

Substituting this in equation $(1)$ ,

$ω0=Fav(h−r)Δt25mr2$ $(2)$

Applying impulse-momentum theorem to the ball,

$I=FavΔt=Δp=mv0$

$⇒mv0=FavΔt$

$v0=FavΔtm$ $(3)$

Where, $m$ is mass of the ball

$Δp$ is momentum, and

$v0$ is speed of the ball just after the impact

From equation $(3)$ , $v0=FavΔtm⇒Δt=mv0Fav$

Substituting the expression for $Δt$ in equation $(3)$ ,

$ω0=Fav(h−R)mv0Fav25mR2=5v0(h−R)2R2$

Substituting $h=2R3$ in the above equation,

$ω0=5v0(2R3−R)2R2=5v0(2R−3R3)2R2=5v0(−R3)2R2=−5v06R$

Conclusion:

The angular speed of the ball just after the blow is $−5v06R$ .

Answer 7

Chapter 9, Problem 106P

(a)

To determine

The angular speed of the ball just after the blow.

Answer 8

(a)

Expert Solution

Answer to Problem 106P

$ω0=−5v06R$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline

Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

Newton’s second law of motion in rotational form

$∑τ=r×F=Icmα$

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 1

FIGURE: 1

Applying Newton’s second law in rotational form to ball,

$∑τ0=Fav(h−R)=Icmα=Icmω0Δt$

$⇒Fav(h−R)=Icmω0Δt$

$ω0=Fav(h−R)ΔtIcm$ $(1)$

Where,

$τ0$ is the torque about the center of the ball

$Fav$ is the average force on the ball

$h$ is the below the centerline at which the force is applied on the ball

$R$ is radius of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$ω0$ is angular speed of the ball after impact

$Δt$ is elapsed time

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mr2$

Substituting this in equation $(1)$ ,

$ω0=Fav(h−r)Δt25mr2$ $(2)$

Applying impulse-momentum theorem to the ball,

$I=FavΔt=Δp=mv0$

$⇒mv0=FavΔt$

$v0=FavΔtm$ $(3)$

Where, $m$ is mass of the ball

$Δp$ is momentum, and

$v0$ is speed of the ball just after the impact

From equation $(3)$ , $v0=FavΔtm⇒Δt=mv0Fav$

Substituting the expression for $Δt$ in equation $(3)$ ,

$ω0=Fav(h−R)mv0Fav25mR2=5v0(h−R)2R2$

Substituting $h=2R3$ in the above equation,

$ω0=5v0(2R3−R)2R2=5v0(2R−3R3)2R2=5v0(−R3)2R2=−5v06R$

Conclusion:

The angular speed of the ball just after the blow is $−5v06R$ .

Answer 13

(b)

To determine

The speed of the ball once it begins to roll without slipping.

(b)

Expert Solution

Answer to Problem 106P

$v=521v0$

Explanation of Solution

Given: Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 2

FIGURE: 2

Referring to the force diagram shown in figure 2, applying Newton’s second law to the ball when it is rolling without slipping,

$Στ0=fkR=Icmα$ $(4)$

$ΣFy=Fn−mg=0$ $(5)$

And

$ΣFx=−fk=ma$ $(6)$

Where,

$Fx$ is force in the x-direction

$fk$ is force of friction in x-direction

$Fy$ is force in the y-direction

$fn$ is force of friction in y-direction

$m$ is mass of the ball

$a$ is the acceleration of the ball

$g$ is acceleration due to gravity

$R$ is radius of the ball

$τ0$ is torque about the center of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$g$ is acceleration due to gravity, $g=9.8 m/s2$

But, $fk=μkFn$ $(7)$

Where, $μk$ is coefficient of kinetic friction

Calculations:

From equation $(5)$ , $Fn=mg$

Substituting this in equation $(7)$ ,

$fk=μkmg$

From equation $(4)$ angular acceleration,

$α=fkRIcm$

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mR2$

Substituting for $fk$ and $Icm$ ,

$α=μkmgR25mR2=5μkg2R$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

$ω=ω0+αΔt=ω0+5μkg2RΔt$ $(8)$

Now,substituting the expression for $fk$ in equation $(6)$ ,

$−μkmg=ma⇒a=−μkg$ $(9)$

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

$v=v0+aΔt$ $(10)$

Where,

$v$ is speed of the ball

$v0$ is speed of the ball just after impact

$a$ is the acceleration ball

$Δt$ is elapsed time

Substituting for $a$ from equation $(9)$ in equation $(10)$ ,

$v=v0−μkgΔt$ $(11)$

Imposing the condition for rolling the ball without slipping,

$R(ω0+5μkg2RΔt)=v0−μkgΔt⇒Δt=1621v0μkg$

Substituting this $Δt$ in equation $(11)$ ,

$v=v0−μkg(1621v0μkg)=521v0$

Conclusion:

The speed of the ball once it begins to roll without slipping is $521v0$ .

Answer 14

(b)

To determine

The speed of the ball once it begins to roll without slipping.

Answer 15

(b)

Expert Solution

Answer to Problem 106P

$v=521v0$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given: Speed of the ball just after the blow is $v0$

The coefficient of kinetic friction between the ball and billiard table is $μk$

Formula Used:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 9, Problem 106P , additional homework tip 2

FIGURE: 2

Referring to the force diagram shown in figure 2, applying Newton’s second law to the ball when it is rolling without slipping,

$Στ0=fkR=Icmα$ $(4)$

$ΣFy=Fn−mg=0$ $(5)$

And

$ΣFx=−fk=ma$ $(6)$

Where,

$Fx$ is force in the x-direction

$fk$ is force of friction in x-direction

$Fy$ is force in the y-direction

$fn$ is force of friction in y-direction

$m$ is mass of the ball

$a$ is the acceleration of the ball

$g$ is acceleration due to gravity

$R$ is radius of the ball

$τ0$ is torque about the center of the ball

$Icm$ is the moment of inertia with respect to an axis through the center of mass of the ball

$α$ is angular acceleration of the ball

$g$ is acceleration due to gravity, $g=9.8 m/s2$

But, $fk=μkFn$ $(7)$

Where, $μk$ is coefficient of kinetic friction

Calculations:

From equation $(5)$ , $Fn=mg$

Substituting this in equation $(7)$ ,

$fk=μkmg$

From equation $(4)$ angular acceleration,

$α=fkRIcm$

Moment of inertia with respect to an axis through the center of mass of the ball is

$Icm=25mR2$

Substituting for $fk$ and $Icm$ ,

$α=μkmgR25mR2=5μkg2R$

Now let us write constant-acceleration equation that connects angular speed of the ball to the angular acceleration and time,

$ω=ω0+αΔt=ω0+5μkg2RΔt$ $(8)$

Now,substituting the expression for $fk$ in equation $(6)$ ,

$−μkmg=ma⇒a=−μkg$ $(9)$

Constant acceleration equation that relates the speed of the ball to the acceleration and time,

$v=v0+aΔt$ $(10)$

Where,

$v$ is speed of the ball

$v0$ is speed of the ball just after impact

$a$ is the acceleration ball

$Δt$ is elapsed time

Substituting for $a$ from equation $(9)$ in equation $(10)$ ,

$v=v0−μkgΔt$ $(11)$

Imposing the condition for rolling the ball without slipping,

$R(ω0+5μkg2RΔt)=v0−μkgΔt⇒Δt=1621v0μkg$

Substituting this $Δt$ in equation $(11)$ ,

$v=v0−μkg(1621v0μkg)=521v0$

Conclusion:

The speed of the ball once it begins to roll without slipping is $521v0$ .

Answer 20

(c)

To determine

The kinetic energy of the ball just after the hit.

(c)

Expert Solution

Answer to Problem 106P

$Ki=23mv0236$

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline.

Speed of the ball just after the blow is $v0$ .

The coefficient of kinetic friction between the ball and billiard table is $μk$ .

Formula Used:

Initial kinetic energy of the ball can be written as,

$Ki=Ktrans+Krot$ $(12)$

Where, $Ktrans$ is kinetic energy due to translational motion of the ball, $Ktrans=12mv02$

$Krot$ is kinetic energy due to rotational energy of the ball, $Krot=12Iω02$

Substituting the expressions for $Ktrans$ and $Krot$ in equation $(12)$ ,

$Ki=12mv02+12Iω02$ $(13)$

Where, $m$ is mass of the ball,

$v0$ is speed of the ball just after impact

$I$ is moment of inertia of the ball

$ω0$ is angular speed of the ball

Moment of inertia, $I=25mR2$ , and Angular speed, $ω0=−5v06R$

Substituting these in equation $(13)$ ,

$Ki=12mv02+12(25mR2)(−5v06R)2$

$=12mv02+15mR2×2536v02R2$

$=12mv02+536mv02$

$=18mv02+5mv0236=23mv0236$

Conclusion:

The kinetic energy of the ball just after the hit is $23mv0236$ .

Answer 21

(c)

To determine

The kinetic energy of the ball just after the hit.

Answer 22

(c)

Expert Solution

Answer to Problem 106P

$Ki=23mv0236$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given: The horizontal force applied on a billiard ball at a distance $h=2R3$ , below the centerline.

Speed of the ball just after the blow is $v0$ .

The coefficient of kinetic friction between the ball and billiard table is $μk$ .

Formula Used:

Initial kinetic energy of the ball can be written as,

$Ki=Ktrans+Krot$ $(12)$

Where, $Ktrans$ is kinetic energy due to translational motion of the ball, $Ktrans=12mv02$

$Krot$ is kinetic energy due to rotational energy of the ball, $Krot=12Iω02$

Substituting the expressions for $Ktrans$ and $Krot$ in equation $(12)$ ,

$Ki=12mv02+12Iω02$ $(13)$

Where, $m$ is mass of the ball,

$v0$ is speed of the ball just after impact

$I$ is moment of inertia of the ball

$ω0$ is angular speed of the ball

Moment of inertia, $I=25mR2$ , and Angular speed, $ω0=−5v06R$

Substituting these in equation $(13)$ ,

$Ki=12mv02+12(25mR2)(−5v06R)2$

$=12mv02+15mR2×2536v02R2$

$=12mv02+536mv02$

$=18mv02+5mv0236=23mv0236$

Conclusion:

The kinetic energy of the ball just after the hit is $23mv0236$ .

Answer 27

Ch. 9 - Prob. 1P Ch. 9 - Prob. 2P Ch. 9 - Prob. 3P Ch. 9 - Prob. 4P Ch. 9 - Prob. 5P Ch. 9 - Prob. 6P Ch. 9 - Prob. 7P Ch. 9 - Prob. 8P Ch. 9 - Prob. 9P Ch. 9 - Prob. 10P

The angular speed of the ball just after the blow.

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The angular speed of the ball just after the blow.

Answer to Problem 106P

Explanation of Solution

The speed of the ball once it begins to roll without slipping.

Answer to Problem 106P

Explanation of Solution

The kinetic energy of the ball just after the hit.

Answer to Problem 106P

Explanation of Solution

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