Find the kinetic energy of the merry-goround after 3.59 s. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder.Answer in units of J.

Question

A horizontal 873 N merry-go-round of radius 1.11 m is started from rest by a constant horizontal force of 70.6 N applied tangentially to the merry-go-round.

Answer 1

One way to go about calculating the final KE is to first compute the final angular momentum, L. The applied torque N is constant,

N = F R

where F = applied force = 70.6N,

and R = radius = 1.11m.

L = N T = F R T

where T = time = 3.59s.

Another expression for final angular momentum L is

L = I ω

where I = moment of inertia of solid cylinder = M R² / 2

M = mass of solid cylinder = weight / g

So M R² ω / 2 = F R T

allows us to calculate the final angular velocity ω. Once that is found, the final kinetic energy is

K = I ω² / 2.

Substituting expressions for I and ω in the last expression for K, some simplification can be achieved. However, all quantities on the right side will be known so the final K can be computed.

Answer 2

I will set this up for you, but YOU do the calculations.

K = kinetic energy

ω = angular speed

I = moment of inertia

F = force

τ = torque = RF

α = angular acceleration

t = time

K = Iω²/2

We need to find I, α, and ω.

For a cylinder,

I = MR²/2

You know the radius R. Get M by dividing the weight of the cylinder by g. Once you have M and R, you can calculate I.

ω(t) = ω₀ + αt = 0 + αt = αt

α = τ/I = RF/I

You know R and F, and you can calculate I from the steps above. Putting it all into the equation for kinetic energy, we get

K = Iω²/2 = I(αt)²/2

You found I and α from the steps above. You know t. Plug in the numbers and finish from here. The answer will come out in joules, given the units of the problem..