Question

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m². The father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 N · m.
(a) Calculate the rotational kinetic energy (in J) in the merry-go-round plus child when they have an angular velocity of 14.8 rpm. J
(b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. revolutions
(c) Again, using energy considerations, calculate the force (in N) the father must exert to stop the merry-go-round in four revolutions. N

Answer 1

Answer:

Part a)

$KE = 101.4 J$

Part b)

$N = 0.043 revolution$

Part c)

F = 2.7 N

Explanation:

Part a)

As we know that the rotational kinetic energy of the merry go round is given as

$KE = \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}84.4(\omega^2)$

here we know that

$\omega = 2\pi(\frac{14.8}{60})$

$\omega = 1.55 rad/s$

Now we have

$KE = \frac{1}{2}(84.4)(1.55^2)$

$KE = 101.4 J$

Part b)

Now we know that work done due to torque = change in kinetic energy

$W = KE_f - KE_i$

$\tau (2N\pi) = 101.4 - 0$

$375(2\pi N) = 101.4$

$N = 0.043 revolution$

Part c)

In order to stop it in four revolutions we have

$\tau(2\pi N) = \Delta KE$

$FR(2\pi N) = 101.4$

$F(1.5)(2\pi \times 4) = 101.4$

F = 2.7 N