A 432 kg merry-go-round in the shape of a horizontal disk with a radius of 2.3 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.1 rad/s in 2.1 s

τ = 1679.68Nm

Explanation:

In order to calculate the required torque you first take into account the following formula:

          (1)

τ: torque

I: moment of inertia of the merry-go-round

α: angular acceleration

Next, you use the following formulas for the calculation of the angular acceleration and the moment of inertia:

        (2)

          (3)       (it is considered that the merry-go-round is a disk)

w: final angular speed = 3.1 rad/s

wo: initial angular speed = 0 rad/s

M: mass of the merry-go-round = 432 kg

R: radius of the merry-go-round = 2.3m

You solve the equation (2) for α. Furthermore you calculate the moment of inertia:

Finally, you replace the values of the moment of inertia and angular acceleration in the equation (1):

The required torque is 1679.68Nm

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.