EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100654428
Author: Jewett
Publisher: Cengage Learning US
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Question
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Chapter 11, Problem 11.48AP

(a)

To determine

The speed at the bottom of the half-pipe.

(a)

Expert Solution
Check Mark

Answer to Problem 11.48AP

The speed at the bottom of the half-pipe is 11.1m/s .

Explanation of Solution

Given info: The mass of particle is 76.0kg , the distance from the ground of its center of mass is 0.500m . The radius of the half of the pipe is 6.80m .

Write the expression for conservation of work and energy law.

KE=U (1)

Here,

KE is the kinetic energy which is at point B.

U is the potential energy when the skateboarder is at rest at point A.

The skateboarder is at rest at point A, so there is a potential energy at point A,

U=mgx

Here,

x is the radius of the circle.

m is the mass of the particle

g is the acceleration due to gravity.

The center of mass moves through one quarter of the circle.

The radius of the circle is,

x=6.80m0.500m=6.30m

The skateboarder is in motion so it acquires the kinetic energy at point B,

KE=12mv2

Here,

v is the speed of skateboarder at the bottom of the half pipe.

Substitute 12mv2 for KE and mgx for U in equation (1).

12mv2=mgxv2=2gxv=2gx

Substitute 6.30m for x and 9.8m/s2 for g in above equation.

v=2(9.8m/s2)(6.30m)=11.1m/s

Conclusion:

Therefore, the speed at the bottom of the half-pipe is 11.1m/s .

(b)

To determine

The angular momentum of him about the center of curvature at the point B.

(b)

Expert Solution
Check Mark

Answer to Problem 11.48AP

The angular momentum of him about the center of curvature at the point B is 5.314×103kgm2/s .

Explanation of Solution

Given info: The mass of particle is 76.0kg , the distance from the ground of its center of mass is 0.500m . The radius of the half of the pipe is 6.80m .

Write the expression for the angular momentum about the center of curvature.

L=mvx

Here,

L is the angular momentum about the center of curvature.

Substitute 6.30m for x , 76.0kg for m and 11.1m/s for v in above equation.

L=(76.0kg)(11.1m/s)(6.30m)=5.314×103kgm2/s

Conclusion:

Therefore, the angular momentum of him about the center of curvature at the point B is 5.314×103kgm2/s .

(c)

To determine

To explain: The angular momentum of him is constant in this maneuver, whereas the kinetic energy of his body is not constant.

(c)

Expert Solution
Check Mark

Answer to Problem 11.48AP

After the passing point B, there is no torque about the axis of the channel act on him so; the angular momentum will be constant, but his legs convert the chemical energy into mechanical energy and the kinetic energy of his body is not constant.

Explanation of Solution

Given info: The mass of particle is 76.0kg , the distance from the ground of its center of mass is 0.500m . The radius of the half of the pipe is 6.80m .

A skateboarder passes the point B, so there is no tangential force acts on him because the wheels on the skate prevent this force. The torque is zero due to no tangential force, so the angular momentum will be constant.

The kinetic energy increase because his legs convert chemical energy into mechanical energy and the kinetic energy will not be constant. While the normal force rises trajectory to enhance his linear momentum.

Conclusion:

Therefore, after the passing point B, there is no torque about the axis of the channel act on him so; the angular momentum will be constant, but his legs convert the chemical energy into mechanical energy and the kinetic energy of his body is not constant.

(d)

To determine

The speed immediately after the skateboarder stands up.

(d)

Expert Solution
Check Mark

Answer to Problem 11.48AP

The speed of skateboarder after he stands up is 12.0m/s .

Explanation of Solution

Given info: The mass of particle is 76.0kg , the distance from the ground of its center of mass is 0.500m . The radius of the half of the pipe is 6.80m .

The skateboarder stands up, so the distance is,

x=6.8m0.950m=5.85m

Write the expression for angular momentum.

L=mvx

Here,

v is the speed of skateboarder at point C.

Substitute 76.0kg for m , 5.85m for x and 5.314×103kgm2/s for L in above equation.

5.314×103kgm2/s=(76.0kg)v(5.85m)v=5.314×103kgm2/s(76.0kg)(5.85m)=11.95m/s12.0m/s

Conclusion:

Therefore, the speed of skateboarder after he stands up is 12.0m/s .

(e)

To determine

The amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up.

(e)

Expert Solution
Check Mark

Answer to Problem 11.48AP

The amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up is 1.125kJ .

Explanation of Solution

Given info: The mass of particle is 76.0kg , the distance from the ground of its center of mass is 0.500m . The radius of the half of the pipe is 6.80m .

At point B, the skate boarder has kinetic and chemical energy is,

Eb=KEB+W

Here,

Eb is the total energy at point B.

W is the chemical; energy.

KEB is the kinetic energy at point B.

At point C, he has kinetic energy due and the potential energy is,

Ec=KEc+Uc

Here,

Ec is the total energy at point C.

KEc is the kinetic energy at point C.

Uc is the potential energy at point C.

Write the expression of the conservation of energy.

Eb=Ec

Substitute KEB+W for Eb and KEc+Uc for Ec in above equation.

KEB+W=KEc+Uc . (2)

Write the expression for the kinetic energy at point B.

KEB=12mv2

Substitute 76.0kg for m and 11.1m/s for v in above equation.

KEB=12(76.0kg)(11.1m/s)2=4681.98kg(m/s)2

Thus, the kinetic energy at point B is 4681.98kg(m/s)2 .

Write the expression for the kinetic energy at point C.

KEc=12mv2

Substitute 76.0kg for m and 12.0m/s for v in above equation.

KEc=12(76.0kg)(12.0m/s)2=5472kg(m/s)2

Thus, the kinetic energy at point C is 5472kg(m/s)2 .

Write the expression for potential energy at point C.

Uc=mgx . (3)

Here,

x is the radius of the pipe at point C.

The radius of the pipe at point C,

x=0.950m0.500m=0.450m

Substitute 0.450m for x , 76.0kg for m and 9.8m/s2 for g in equation (3).

Uc=(76.0kg)(9.8m/s2)(0.450m)=335.16kg(m/s)2

Thus, the potential energy at point C is 335.16kg(m/s)2 .

Substitute 335.16kg(m/s)2 for Uc , 5472kg(m/s)2 for KEc and 4681.98kg(m/s)2 for KEB in equation (2).

4681.98kg(m/s)2+W=5472kg(m/s)2+335.16kg(m/s)2W=5807.16J4681.98J=1125.18J=1.125kJ

Conclusion:

Therefore, the amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up is 1.125kJ .

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