PHYSICS FOR SCI & ENGR W WEBASSIGN
PHYSICS FOR SCI & ENGR W WEBASSIGN
10th Edition
ISBN: 9781337888486
Author: SERWAY
Publisher: CENGAGE L
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Chapter 10, Problem 33P

A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on a horizontal section of a track as shown in Figure P10.33. It rolls around the inside of a vertical circular loop of radius r = 45.0 cm. As the ball nears the bottom of the loop, the shape of the track deviates from a perfect circle so that the ball leaves the track at a point h = 20.0 cm below the horizontal section. (a) Find the ball’s speed at the top of the loop. (b) Demonstrate that the ball will not fall from the track at the top of the loop. (c) Find the ball’s speed as it leaves the track at the bottom. (d) What If? Suppose that static friction between ball and track were negligible so that the ball slid instead of rolling. Describe the speed of the ball at the top of the loop in this situation. (e) Explain your answer to part (d).

Figure P10.33

Chapter 10, Problem 33P, A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on

(a)

Expert Solution
Check Mark
To determine

The speed of the ball at the top of the loop.

Answer to Problem 33P

The speed of the ball at the top of the loop is 2.38m/s.

Explanation of Solution

The radius of the circular loop is 45.0cm, the height of the track from the horizontal section is 20.0cm and the initial velocity of the ball is 4.03m/s.

Write the expression for the law of energy conservation between horizontal track and the top of the loop

    Kt1+Kr1=Kt2+Kr2+U2        (1)

Here, Kt1 is the translational kinetic energy of the ball at horizontal track, Kr1 is the rotational kinetic energy of the ball at horizontal track, Kt2 is the translational kinetic energy of the ball at top point of the loop, Kr2 is the rotational kinetic energy of the ball at top point of the loop and U2 is the potential energy of the ball at top point of the loop.

Write the expression for the translational kinetic energy of the ball at horizontal track

    Kt1=12mv12

Here, m is the mass of the ball and v1 is the initial linear velocity of the ball.

Write the expression for the rotational kinetic energy of the ball at horizontal track

    Kr1=12Iω12        (2)

Here,  I is the moment of inertia of the ball about the center of mass and ω1 is the initial angular speed of the ball.

Write the expression for the moment of inertia of the ball

    I=23mr2

Write the expression for the initial angular speed of the ball

    ω1=v1r

Substitute 23mr2 for I and v1r for ω1 in equation (2).

    Kr1=12(23mr2)(v1r)2=13mv12

Write the expression for the translational kinetic energy of the ball at top point of the loop

    Kt2=12mv22

Here, v2 is the final linear velocity of the ball.

Write the expression for the rotational kinetic energy of the ball at top point of the loop

    Kr2=12Iω22        (3)

Here, ω2 is the initial angular speed of the ball.

Write the expression for the moment of inertia of the ball

    I=23mr2

Write the expression for the initial angular speed of the ball

    ω2=v2r

Substitute 23mr2 for I and v2r for ω2 in equation (2).

    Kr2=12(23mr2)(v2r)2=13mv22

Write the expression for the potential energy of the ball at top point of the loop

    U2=mg(2r)

Here, g is the acceleration due to gravity.

Substitute 12mv12 for Kt1, 13mv12 for Kr1, 12mv22 for Kt2, 13mv22 for Kr2 and mg(2r) for U2 in equation (1).

    12mv12+13mv12=12mv22+13mv22+mg(2r)

Simplify the above equation for v2.

    56mv22=56mv12mg(2r)v2=v1265g(2r)

Conclusion:

Substitute 9.8m/s2 for g, 45.0cm for r and 4.03m/s for v1 in above equation to find v2.

    v2=(4.03m/s)265×9.8m/s2×(2×45.0cm×102m1cm)=16.24m2/s210.58m2/s2=2.38m/s

Therefore, the speed of the ball at the top of the loop is 2.38m/s.

(b)

Expert Solution
Check Mark
To determine

The reason that the ball will not fall from the track at the top of the loop.

Answer to Problem 33P

The ball will not fall because the value of the centripetal acceleration is more than the acceleration due to gravity at the top point of the circular loop.

Explanation of Solution

The radius of the circular loop is 45.0cm, the height of the track from the horizontal section is 20.0cm and the initial velocity of the ball is 4.03m/s.

Formula to calculate the centripetal acceleration on the ball at the top of the loop

    a=(v2)2r

Here, a is the centripetal acceleration act on the ball at the top of the loop.

Substitute 2.38m/s for v2 and 45.0cm for r to find a.

    a=(2.38m/s)245.0cm×102m1cm=12.58m/s2

Thus, the centripetal acceleration act on the ball at the top of the loop is 12.58m/s2.

Since the centripetal acceleration at the top of the loop is more than the acceleration due to gravity that is (a>g) hence the ball remains on the track at top of the loop and will not fall.

Conclusion:

Therefore, the ball will not fall because the value of the centripetal acceleration is more than the acceleration due to gravity at the top point of the circular loop.

(c)

Expert Solution
Check Mark
To determine

The speed of the ball as it leaves the track at the bottom.

Answer to Problem 33P

The speed of the ball as it leaves the track at the bottom is 4.31m/s.

Explanation of Solution

The radius of the circular loop is 45.0cm, the height of the track from the horizontal section is 20.0cm and the initial velocity of the ball is 4.03m/s.

Write the expression for the law of energy conservation between horizontal track and the bottom of the loop

    Kt1+Kr1=Kt2+Kr2+U2        (4)

Here, Kt2 is the translational kinetic energy of the ball at bottom point of the loop, Kr2 is the rotational kinetic energy of the ball at bottom point of the loop and U2 is the potential energy of the ball at bottom point of the loop.

Write the expression for the translational kinetic energy of the ball at bottom point of the loop

    Kt3=12mv32

Here, v3 is the linear velocity of the ball at bottom.

Write the expression for the rotational kinetic energy of the ball at bottom point of the loop

    Kr3=12Iω32        (5)

Here,  ω3 is the initial angular speed of the ball.

Write the expression for the initial angular speed of the ball

    ω3=v3r

Substitute 23mr2 for I and v3r for ω3 in equation (5).

    Kr3=12(23mr2)(v3r)2=13mv32

Write the expression for the potential energy of the ball at top point of the loop

    U3=mgh

Here, g is the acceleration due to gravity and h is the height of the bottom track from the horizontal section.

Substitute 12mv12 for Kt1, 13mv12 for Kr1, 12mv32 for Kt3, 13mv32 for Kr3 and mgh for U3 in equation (4).

    12mv12+13mv12=12mv32+13mv32mgh

Simplify the above equation for v3.

    56mv32=56mv12+mghv3=v12+65gh

Conclusion:

Substitute 9.8m/s2 for g, 20.0cm for h and 4.03m/s for v1 in above equation to find v3.

    v3=(4.03m/s)2+65×9.8m/s2×20.0cm×102m1cm=16.24m2/s2+2.35m2/s2=4.31m/s

Thus, the speed of the ball as it leaves the track at the bottom is 4.31m/s.

(d)

Expert Solution
Check Mark
To determine

The speed of the ball at the top of the loop if ball slide instead of roll.

Answer to Problem 33P

The speed of the ball at the top of the loop is imaginary.

Explanation of Solution

Write the expression for the law of energy conservation between horizontal track and the top of the loop

    Kt1=Kt2'+U2'        (6)

Here, Kt1 is the translational kinetic energy of the ball at horizontal track, Kt2' is the new translational kinetic energy of the ball at top point of the loop and U2' is the new potential energy of the ball at top point of the loop.

Write the expression for the rotational kinetic energy of the ball at horizontal track

Write the expression for the new translational kinetic energy of the ball at top point of the loop

    Kt2=12m(v2')2

Here, v2' is the new linear velocity of the ball at the top point.

Write the expression for the new potential energy of the ball at top point of the loop

    U2'=mg(2r)

Substitute 12mv12 for Kt1, 12m(v2')2 for Kt2', and mg(2r) for U2 in equation (3).

    12mv12=12m(v2')2+mg(2r)

Simplify the above equation for v2.

    12m(v2')2=11mv12mg(2r)v2'=v122g(2r)

Substitute 9.8m/s2 for g, 45.0cm for r and 4.03m/s for v1 in above equation to find v2'.

    v2'=(4.03m/s)22×9.8m/s2×(2×45.0cm×102m1cm)=16.24m2/s217.64m2/s2=1.4m2/s2

Since the value inside the square root is negative that means the value is imaginary. This condition is impractical.

Thus, the speed of the ball at the top of the loop is imaginary.

Conclusion:

Therefore, the speed of the ball at the top of the loop is imaginary. It can’t be calculated.

(e)

Expert Solution
Check Mark
To determine

The explanation of the solution of part (d).

Answer to Problem 33P

The ball has not sufficient energy to arrive at top of the circular loop.

Explanation of Solution

The velocity comes out to be imaginary in part (d) that indicates the situation in impossible because as the boll slide instead of rolling, the ball have only translational kinetic energy which is insufficient for the ball to reach the top point on the circular loop.

Thus, the ball did not arrive at the top point of the loop.

Conclusion:

Therefore, the ball has not sufficient energy to arrive at top of the circular loop.

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Chapter 10 Solutions

PHYSICS FOR SCI & ENGR W WEBASSIGN

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