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Sushma · Accepted Answer

Rotational Energy Solutions
Q. 1. A 10.0-kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 10.0 m/s, determine 
(a) the translational kinetic energy of its center of gravity
Ans: Kt = 1/2mv2
Here m = 10 Kg, v = 10 m/s so Kt = 500 J
(b) the rotational kinetic energy about its center of gravity
Ans: Kr = 1/2 Iω2 as I = 1/2mr2 and ω = v/r
putting the value of I in equation of Kr we get,
Kr = 1/4mv2
on putting the values m = 10 Kg, v = 10 m/s, we get Kr = 250 J
(c) its total kinetic energy.
Ans: Total K.E. = Kt + Kr = 500 + 250 = 750 J
Q. 2. The net work done in accelerating a propeller from rest to an angular speed of 200 rad/s 
is 3000 J. What is the moment of inertia of the propeller?
Ans: Work done = change in kinetic energy = 1/2I ω2f - 1/2I ω2i       ------[1]
Here, I is the moment of inertia and ω is angular velocity.
Here, ωi = 0 and ωf = 200 rad/s and work = 3000 J
On putting values in equation 1 we get, 
3000 = ½ I 200*200
I = 0.15Kgm2
3.  A horizontal 800-N merry-go-round of radius 1.50 m is started from rest by a constant horizontal force of 50.0 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 3.00 s. (Assume it is a solid cylinder.
Ans: Weight of Merry -go-round = 800 N, Mass of Merry -go-round (m) = W/g =800N/9.8m/s2 = 81.6 Kg, Radius of Merry -go-round (r) =1.50 m, applied force = 50N
Initial angular speed ωi = 0
As we know that Ʈ =Iά and Ʈ = F* r and I = 1/2mr2 
then we calculate ά as: 
ά = Ʈ/I = F* r/1/2mr2 = 2F/mr ------- [1]
putting values in equation we get, 
 ά = 0.817 rad/s2
now, ωf = ωi + άt-------- [2]
at t = 3.0 S, putting value in equation 2 we get, 
ωf = 2.45 rad/s
Now, kinetic energy of the merry-go-round after 3 s will be:
KE = 1/2I ωf2
  Putting I = 1/2mr2, we get KE = ½(1/2mr2) ωf2--------[3]
Putting values in equation 3 we get, KE = 276 J
4. A 240-N sphere 0.20 m in radius rolls, without slipping, 6.

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