Question

a 1.9 solid cylinder (radius =0.20 m, length = 0.60 m) is released from rest at the top of the 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? what is it’s rotational kinetic energy?

Answer 1

We know energy is neither created nor destroyed but converted or transferred from one form to another.

Thus as the cylinder rolls down the ramp all of its potential energy is converted into kinetic energy of the cylinder.And since cylinder has both translational and rotational kinetic energy (because while going down the ramp cylinder has two motion associated with it 1.) rolling about its own axis and 2.)translation along ramp), we need to determine the relationship between the equations for these two types of kinetic energy.
We have,
Translational KE = 1/2 * m * v² ———(1)
Rotational KE = 1/2 * I * ω^{2 ——————(2)}
For a solid cylinder, moment of inertia(I) = 1/2 * m * r²
We also know angular velocity,ω = v/r ==>v²/r²

Therefore from equation 1 and 2 ,Rotational KE = 1/2 * 1/2 * m * r²*v²/r²= 1/4 * m * v^{2 ————(3)}
The cylinder’s total kinetic energy at the bottom of the ramp is equal to the sum of its translational and rotational kinetic energy.

Total KE = translational KE + rotational KE = 1/2 * m * v² + 1/4 * m * v²

Therefore, total KE = 3/4 * m * v²= 3/4 * 1.9 * v²= 1.425 * v² —————(4)
Now for finding cylinder’s velocity at the bottom of the ramp, equate total kinetic energy at bottom of ramp to its initial potential energy at top of the ramp.

Initial PE = mgh =  1.9 * 9.8 * 0.8= 14.896
Therefore, 1.425 * v² = 14.896
v² = 14.896 / 1.425
v = √(14.896/1.425) = 3.23 m/s
Therefore now putting value of v in total kinetic energy equation(4), we get
Total KE = 1.425 (3.23)² = 14.896 J [Answer part a]
Now, cylinder’s rotational kinetic energy use equation(3).

Rotational KE = 1/4 * 1.9 * 3.23² = 4.955 J [Answer part b]