Question

A 1.5 kg solid cylinder (radius = 0.15 m , length = 0.60 m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.80 m high and 5.0 m long.

When the cylinder reaches the bottom of the ramp, what is its total kinetic energy?

When the cylinder reaches the bottom of the ramp, what is its rotational kinetic energy?

When the cylinder reaches the bottom of the ramp, what is its translational kinetic energy?

Answer 1

Gven,

Mass of the cylinder, m = 1.5 kg

radius of the cylinder, r = 0.15 m

lemgth, l = 0.60 m

Height of the ramp, h = 0.80 m

Now,

Let the velocity at the bottom of the ramp be v

Now, applying conservation of energy

=> iniitial energy = final energy

=> mgh = K_f

=> K_f = 1.5*9.8*0.80 = 117.6 J

Total kinetic energy at the bottom is 117.6 J

Now,

Total Kinetic energy, K_f = 1/2*m*v² + 1/2*I*²

AS we know,

= v/r    and I = mr²/2

=> K_f = 1/2*m*v² + 1/2*(mr²/2)*(v/r)²

          = 1/2*m*v² + 1/4*m*v² = 3/4*m*v²

or K_f = 3/4*m*v²

Now,

rotational kinetic energy, K_R = 1/2*I*² = 1/4*m*v²

                                               = 1/3*( 3/4*m*v²) = 1/3*K_f

                                               = 1/3*117.6 = 39.2 J

Transitional kinetic energy, K_T = 1/2*m*v² = 2/3*( 3/4*m*v²)

                                                  = 2/3*K_f = 2/3*117.6

                                                  = 78.4 J