physics kinetic energy?

  • Thread starter Thread starter Grace P
  • Start date Start date
G

Grace P

Guest
A 10.0 kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 11.0 m/s.

(a) Determine the translational kinetic energy of its center of gravity. Joules
(b) Determine the rotational kinetic energy about its center of gravity. Joules
(c) Determine its total kinetic energy. Joules
 
Translational KE = 1/2*m*v^2. Plug in. Done.
Rotational KE = 1/2*I*omega^2. I for a solid cylinder is .5*mr^2. r is not given. So you don't have enough info to complete this part of the problem.
Total KE = Translational + Rotational.
 
Let m = 10 kg and v = 11 mps hub and tangential velocity

a) Let translational KE be KEt = 1/2 mv^2

b) Let rotational KE be KEr = 1/2 kmr^2 w^2 = 1/2 k mv^2. Where w is the angular velocity and v = wr for radius of rotation r. k = 1 or 1/2 depending on if the cylinder is hollow or solid (you failed to specify).

c) Let total kinetic energy be KE = KEr + KEt = 1/2 mv^2(k + 1)

Everything on the RHS of each equation is given; you can do the math. The trick to solving is in recognizing the hub velocity and the tangential velocity are the same magnitude even though they are going in opposite directions. This follows because there is no slippage.
 
Back
Top