A merry-go-round in the park has a radius of 1.5 m and a rotational inertia of 800 kg·m2. A child pushes the merry-go-round with a constant force of 92 N applied at the edge and parallel to the edge. A frictional torque of 14 N·m acts at the axle of the merry-go-round.
- a. What is the net torque acting on the merry-go-round about its axle?
- b. What is the rotational acceleration of the merry-go-round?
- c. At this rate, what will the rotational velocity of the merry-go-round be after 16 s if it starts from rest?
- d. If the child stops pushing after 16 s, the net torque is now due solely to the friction. What then is the rotational acceleration of the merry-go-round? How long will it take for the merry-go-round to stop turning?
(a)
The net torque acting on the merry –go-round about its axle.
Answer to Problem 1SP
The net torque acting on the merry-go-round is
Explanation of Solution
Given info: Radius is
Write the expression for the net torque.
Here,
Substitute
The net torque acting on the merry-go round is given by subtracting from the frictional torque.
Conclusion:
Therefore, the net torque acting on the merry-go-round is
(b)
The rotational acceleration of the merry-go-round.
Answer to Problem 1SP
The rotational acceleration of the merry-go-round is
Explanation of Solution
Write the expression for the rotational acceleration.
Here,
Substitute
Conclusion:
Therefore, the rotational acceleration of the merry-go-round is
(c)
The rotational velocity of the merry-go-round after
Answer to Problem 1SP
The rotational velocity of the merry-go-round will be
Explanation of Solution
Write the expression for the rotational velocity.
Here,
Substitute
Conclusion:
Therefore, the rotational velocity of the merry-go-round will be
(d)
Rotational acceleration of the merry-go-round and the time taken for the merry-go-round to stop running.
Answer to Problem 1SP
The rotational acceleration will be
Explanation of Solution
The net torque is only due to the friction.
Write the expression for the rotational acceleration.
Substitute
Write the expression for the rotational velocity.
Rearrange the above equation to find
Substitute
Conclusion:
Therefore, rotational acceleration will be
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Chapter 8 Solutions
Physics of Everyday Phenomena
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