Question

A horizontal 800 N merry go round is a solid disk of radius 1.50 m, started from rest by a constant horizontal force of 50.0 N applied tangentially to the edge of the disk. Find the kinetic energy of the disk after 3.00 s.

Answer 1

Answer:

275.6 J

Explanation:

Weight, mg = 800 N

m = 800 / 9.8 = 81.63 kg

R = 1.5 m

F = 50 N

torque = I x α = F x R

where, I be the moment of inertia

α = F x R / I

now, ω = ωo + α t

ω = 0 + F x R x t / I

Kinetic energy, K = 1/2 x I x ω^2

K = 0.5 x (F^2 x R^2 x t^2) / I

K = 0.5 x 50 x 50 x 1.5 x 1.5 x 3 x 3 / (0.5 x 81.63 x 1.5 x 1.5)

K = 275.6 J

Answer 2

The kinetic energy of the disk after 3 seconds will be $KE= 275.6\ J$

What will be the Kinetic energy of the disk after 3 seconds?

It is given that

Weight, mg = 800 N

Then mass will be

$m=\dfrac{800}{9.81}=81.63kg$

The radius of the disk R = 1.5 m

Force applied tangentially to the disk F = 50 N

Now we know that

$Torque =I\times \alpha =F\timesR$

$I=mr^2$

Here,

I will be the moment of inertia

$\alpha =\dfrac{F\times R}{I}$

now,

$w=w_0+\alpha t$

$w=0+(\dfrac{R\times F\times T}{I})$

now Kinetic energy

$KE= \frac{1}{2} \times I\times w^2$  

$KE =0.5\times I(\dfrac{F^2\times R^2\times T^2)}{I^2}$

$KE= 0.5\times (\dfrac{50\times 50 \times1.5\times 1.5\times 3\times 3}{(0.5\times81.63\times1.5\times)} )$

$K=275.6\ J$

Thus the kinetic energy of the disk after 3 seconds will be $KE= 275.6\ J$

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