A merry-go-round, having a radius of 2.5 m, is set in motion starting from rest, by students applying a force of 500 N tangential to the rim of the wheel. After five seconds of torque, the students let go and measure the rotational speed: 1.25 rad/s. The rotational inertia of the merry-go-round is A) 7.5 $\times$ 10² kg m². B) 1.7 $\times$ 10³ kg m². C) 2.5 $\times$ 10³ kg m². D) 5.0 $\times$ 10³ kg m². E) 1.0 $\times$ 10⁴ kg m².

Question

Quizplus · Accepted Answer

The rotational inertia (I) of the merry-go-round can be calculated using the equation:

I = (torque)/(angular acceleration)

The torque applied by the students is 500 N (the force) multiplied by the radius of the merry-go-round (2.5 m):

torque = 500 N * 2.5 m = 1250 Nm

The angular acceleration can be calculated using the equation:

angular acceleration = (change in angular velocity)/(time)

The initial angular velocity is zero (since the merry-go-round starts from rest), and the final angular velocity is 1.25 rad/s. The time is 5 seconds.

angular acceleration = (1.25 rad/s - 0 rad/s) / 5 s = 0.25 rad/s^2

Now we can plug in the values to find the rotational inertia:

I = 1250 Nm / 0.25 rad/s^2 = 5000 kg m^2

Therefore, the correct answer is D: 5.0 x 10^3 kg m^2.

A Merry-Go-Round, Having a Radius of 2 ×\times× 102 Kg M2 ×\times×

A Merry-Go-Round, Having a Radius of 2 $\times$ 10² Kg M² $\times$