COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
COLLEGE PHYSICS LL W/ 6 MONTH ACCESS
2nd Edition
ISBN: 9781319414597
Author: Freedman
Publisher: MAC HIGHER
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Chapter 8, Problem 70QAP
To determine

Normal force acting on the billiard ball at point B

Expert Solution & Answer
Check Mark

Answer to Problem 70QAP

  FB=1.143N

Explanation of Solution

Given info:

Billiard ball of,

  Radius =r0=2.50cm

  Mass =mo=160.0g

Translational speed of billiard ball at A = 2.00ms1 =v0

Point B is at the top of a hill that has a radius of curvature of 60cm

  (r)

Billiard ball rolls without slipping down the track.

Formula used:

Let's name normal force acting on the ball at point B

As FB.

Let's name the angular velocity of billiard ball at the top of the track as ω.

Let's name the linear speed of billiard ball at the top of the hill as vs

Let's name the moment of inertia of billiard ball as Is.

Let's name the vertical distance from potential energy zero level =h =10cm

  g=10ms2.

Conservation of mechanical energy:

  Ki+Ui=Kf+Uf

Kinetic energy for an object that undergoes both translation and rotation:

  K=12Mv2s+12Isω2...(1)

Condition for rolling without slipping:

  vs=roω...(2)

Centrifugal power of billiard ball,

  FB=movs2r

Calculation:

Let's consider the motion of sphere,

Initially the billiard ball is at rest with translational kinetic energy, so Ki=12m0v02.

The initial gravitational potential energy is Ui=mog(0)=0

Final gravitational potential energy is Uf=mog(h)

Conservation of mechanical energy:

  Ki+Ui=Kf+Uf

  12m0v02+(0)=Kf+mog(h)

  Kf=12movo2mogh...(A)

Let's consider the kinetic energy (Kf) of billiard ball at the top of the hill,

Kinetic energy is part translational and part rotational. We can use (2) equation to write ω

In terms of vs.

Using (1) expression,

Kinetic energy for an object that undergoes both translation and rotation:

  K=12Mv2s+12Isω2...(1)

  Kf=12mov2s+12Isω2

Condition for rolling without slipping:

  vs=roω...(2)

  ω=vsro

Substitute into kinetic energy equation:

  Kf=12mov2c+12Icω2

  Kf=12mov2s+12Is( v s r o )2

  Kf=12(mo+Isr02)vs2

From the general knowledge we know that moment of inertia of a sphere is 25m0ro2.

So, let's substitute the Is value in to the equation,

  Kf=12(mo+Isr02)vs2

  Kf=12(mo+( 2 5 m 0 r o 2 )ro2)vs2

  Kf=12(mo+25m0)vs2

  Kf=12(75m0)vs2

  Kf=710movs2...(B)

Since (A),(B) equations are equal,

  (A)=(B)

  12m0v02mog(h)=710movs2

  12v02g(h)=710vs2

  107(12v02g(h))=vs2...(3)

Let's consider the billiard ball at the top of the hill,

Centrifugal power of billiard ball,

  FB=movs2r

Substituting vs2 value to above equation,

  FB=movs2r

  FB=mo107(12v02g(h))r

  FB=mo10(v022g(h))14r

Let's substitute the values,

  FB=mo10(v022g(h))14r

  FB=mo10( (2.00m s 1 )22*10ms 2*(10* 10 2m))14r

  FB=mo10((4.00m2s 2+2.00m2s 2)14r

  FB=160*103kg*10(6.00m2s 2)14*60*102m

  FB=1614N

  FB=1.143N

Conclusion:

Thus, normal force acting on the billiard ball at point B is 1.143N.

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Chapter 8 Solutions

COLLEGE PHYSICS LL W/ 6 MONTH ACCESS

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