A solid cylinder rolls without slipping on a horizontal surface such that its center of mass moves at 5.0 m/s. If it has a total kinetic energy is 1.7 kJ, what is its mass?

Question

anonymous · Answer

Assuming that by not slipping the cylinder in experiencing so little loss of velocity while the velocity is being measured and that it is staying intact (no loss of mass) we can use the expression

K.E = !/2 (mass)(velocity)^2

K.E.            = 1.7 * 10^3 Joules
velocity      = 5.0 m/s
mass          = ?

KE = 1/2 Mv^2.... substitute known values and rearrange to calculate the unknown mass

1.7*10^3  = (0.5)M (5)^2    (I've left out the dimensions...I'm new to using this)

M = ( (2) * (1.7*10^3) ) / 25 Kg   ( I don't have a calculator to hand either :( )

anonymous · Answer

The body is rolling, so it has both rotational and translational KE. 

KE = 1/2 mv^2 + 1/2 Iw^2

Since rolling without slipping, 

v = rw

For solid cylinder 

I = 1/2 mr^2

=> Iw^2 =1/2 mv^2

So KE = 3/4 mv^2 = 1700 = 3/4 m 5^2 

m= (1700x4)/75 = 90.7Kg

anonymous · Answer

good point !

anonymous · Answer

@OdysseusBohs: Oi! Your answer is wrong! What about the rotational KE?