Question

A uniform, solid cylinder of radius 5.00 cm and mass 3.00 kg starts from rest at the top of an inclined plane that is 2.00 m long and tilted at an angle of 25.0° with the horizontal and rolls without slipping down the ramp. What is the cylinder's speedat the bottom of the ramp?

Answer 1

To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is

KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f

$KE_i = 0$, since initially cylinder was at rest

$PE_f = 0$ since at the ground potential energy is zero

The mathematical values are,

$mgh = \frac{1}{2} mV^2 + \frac{1}{2}I\omega^2$

Here,

m = mass

g= Gravity

h = Height

V = Velocity

$I = \frac{mr^2}{2}$ moment of Inertia in terms of its mass and radius

$\omega = \frac{V}{r}$ Angular velocity in terms of tangential velocity and its radius

Replacing the values we have that

mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2

gh = \frac{v^2}{2}+\frac{v^2}{4}

v = \sqrt{\frac{4gh}{3}}

From trigonometry the vertical height of inclined plane is the length of this plane for $sin\theta$, then

$h = 2.00*sin 25$

$h = 0.845 m$

Replacing,

$v = \sqrt{\frac{4(9.8)(0.845)}{3}}$

$V = 3.32 m/s$

Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s