A solid cylindrical disk whose mass is 1.4 kg and radius is 8.5 cm, rolls across a horizontal table at a speed of 15 cm/...

Given:

The mass of the cylinder is 1.4 kg.

The radius of the cylinder is 8.5 cm.

The speed of the cylinder is 15 cm/s.

 

 

Calculation:

Part-I:

Write the expression for the angular speed.

$\omega =\frac{v_{cm}}{R}$

Here, $\omega$ is the angular speed, R is the radius, and v_cm is the speed of the center of the mass.

Substitute, 15 cm/s for v_cm, and 8.5 cm for R in the above expression.

$\begin{array}{rcl}\omega & =& \frac{15 \text{cm/s}}{8.5 \text{cm}}\\ & =& 1.765 \text{rad/s}\\ & & \end{array}$

Thus, the angular speed is 1.765 rad/s.

Part-II:

Write the expression for the instantaneous velocity of the top of the cylinder.

$v=v_{cm}+R\omega$

Here, v is the instantaneous velocity of the top of the cylinder.

Substitute, 1.765 rad/s for $\omega$, 8.5 cm for R, and 15 cm/s for v_cm, in the above expression.

$\begin{array}{rcl}v& =& \left(15\text{ cm/s}\right)+\left(8.5\text{ cm}\right)\left(1.765 \text{rad/s}\right)\\ & =& 30\text{ cm/s}\end{array}$

Thus, the instantaneous velocity of the top edge of the cylinder is 30 cm/s.

Part-III:

Write the expression for the total kinetic energy of the cylinder.

total kinetic energy = rotational kinetic energy + translational kinetic energy.

$E_k=\frac{3}{4}m{v_{cm}}^2$

Substitute, 15 cm/s for v_cm, and 1.4 kg for m in the above expression.

$\begin{array}{rcl}{E}_k& =& \frac{3}{4}\left(1.4\text{ kg}\right){\left\{\left(15 \text{cm/s}\right)\left(\frac{0.01\text{ m/s}}{1\text{ cm/s}}\right)\right\}}^2\\ & =& 0.023625\text{ J}\end{array}$

Thus, the kinetic energy of the cylinder is 0.023525 J.