Ahorizontal 745 n merry-go-round of radius
1.45 m is started from rest by a constant
horizontal force of 56.3 n applied tangentially
to the merry-go-round.
find the kinetic energy of the merry-goround after 3.62 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.

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =

Substituting the values

Moment of Inertia of solid cylinder I  

=>

=>

=>

Step 3: Finding the Torque applied T

Torque applied T =

Substituting the values

T =

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,

Substituting the values,

 Step 4: Finding the Final angular velocity

Final angular velocity ,

Substituting the values,

Now KE (100% rotational) after 3.62s is:

KE =

KE =

KE = 544J

i think the answer would be reserve.

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