Question

A solid, uniform disk of radius 0.250 m and mass 45.2 kg rolls down a ramp of length 5.40 m that makes an angle of 17.0° with the horizontal. The disk starts from rest from the top of the ramp.
(a) Find the speed of the disk's center of mass when it reaches the bottom of the ramp. m/s
(b) Find the angular speed of the disk at the bottom of the ramp. rad/s

Answer 1

Explanation:

Given that,

Radius of the disk, r = 0.25 m

Mass, m = 45.2 kg

Length of the ramp, l = 5.4 m

Angle made by the ramp with horizontal, $\theta=17^{\circ}$

Solution,

As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :

(a) $mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

h is the height of the ramp

$sin\theta=\dfrac{h}{5.4}$

$h=sin(17)\times 5.4=1.57\ m$

v is the speed of the disk's center

I is the moment of inertia of the disk,

$I=\dfrac{1}{2}mr^2$

$\omega=\dfrac{v}{r}$

$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\dfrac{1}{2}mr^2\times (\dfrac{v}{r})^2$

$gh=\dfrac{1}{2}v^2+\dfrac{1}{4}v^2$

$gh=\dfrac{3}{4}v^2$

$9.8\times 1.57=\dfrac{3}{4}v^2$

v = 4.52 m/s

(b) At the bottom of the ramp, the angular speed of the disk is given by :

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{4.52}{0.25}$

$\omega=18.08\ rad/s$

Hence, this is the required solution.