Question

A cylinder with moment of inertia I about its center of mass, mass m, and radius r has a string wrapped around it which is tied to the ceiling (Figure 1) . The cylinder's vertical position as a function of time is y(t).

At time t=0 the cylinder is released from rest at a height h above the ground.

Part B

In similar problems involving rotating bodies, you will often also need the relationship between angular acceleration, ?, and linear acceleration, a. Find ? in terms of a and r.

?=?

Part C

Suppose that at a certain instant the velocity of the cylinder is v. What is its total kinetic energy, Ktotal, at that instant?

Express Ktotal in terms of m, r, I, and v.  

  

Part D

Find vf, the cylinder's vertical velocity when it hits the ground.

Express vf, in terms of g, h, I, m, and r.

?

Answer 1

Concepts and reason

The concept required to solve this problem is conservation of energy in circular motion.

Initially, use the expression of velocity in terms of angular velocity to solve for the relation between the linear acceleration and angular acceleration.

Later, use the equation of total kinetic energy and relation between angular and linear velocity to solve for the expression of total kinetic energy.

Finally, use the law of conservation of energy to solve for the final velocity of the cylinder when it hits the ground.

Fundamentals

The relation between the linear and angular velocity is,

$v = r\omega$

Here, $v$ is the linear speed, $r$ is the radius of the circular motion, and $\omega$ is the angular velocity.

The angular acceleration is,

$\alpha = \frac{{d\omega }}{{dt}}$

Here, $\omega$ is the angular velocity at time $t$.

The linear acceleration of the object is,

$a = \frac{{dv}}{{dt}}$

Here, $v$ is the linear speed at any time $t$.

The total kinetic energy of the cylinder is,

${K_{\rm{T}}} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$

Here, $m$ is the mass, $v$ is the linear velocity, $I$ is the moment of inertia, and $\omega$ is the angular velocity of the object.

The potential energy of the cylinder is,

$U = mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the height it falls.

The expression of conservation of energy gives the change in the total kinetic energy is equal to the negative of change in potential energy.

$\Delta {K_{\rm{T}}} = - \Delta U$

Here, $\Delta {K_{\rm{T}}}$ is the change in total kinetic energy, and $\Delta U$ is the potential energy.

(B)

Use the equation of linear acceleration.

Substitute $r\omega$ for $v$ in the equation $a = \frac{{dv}}{{dt}}$.

$\begin{array}{c}\\a = \frac{{d\left( {r\omega } \right)}}{{dt}}\\\\ = r\frac{{d\omega }}{{dt}}\\\end{array}$

Substitute $\alpha$ for $\frac{{d\omega }}{{dt}}$ in the equation $a = r\frac{{d\omega }}{{dt}}$.

$a = r\alpha$

(C)

The total kinetic energy of the cylinder is,

${K_{\rm{T}}} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$

Here, $m$ is the mass, $v$ is the linear velocity, $I$ is the moment of inertia, and $\omega$ is the angular velocity of the object.

Substitute $\frac{v}{r}$ for $\omega$ in the equation ${K_{\rm{T}}} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$.

${K_{\rm{T}}} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\left( {\frac{v}{r}} \right)^2}$

(D)

Use the equation of conservation of energy.

$\Delta {K_{\rm{T}}} = - \Delta U$

Substitute $\frac{1}{2}mv_{\rm{f}}^2 + \frac{1}{2}I\left( {\frac{{v_{\rm{f}}^2}}{{{r^2}}}} \right)$ for $\Delta {K_{\rm{T}}}$ and $mgh$ for $\Delta U$ in the equation $\Delta {K_{\rm{T}}} = - \Delta U$ and solve for the final velocity ${v_{\rm{f}}}$.

$\begin{array}{c}\\\frac{1}{2}mv_{\rm{f}}^2 + \frac{1}{2}I\left( {\frac{{v_{\rm{f}}^2}}{{{r^2}}}} \right) = - mgh\\\\v_{\rm{f}}^2\left( {m + \frac{I}{{{r^2}}}} \right) = - 2mgh\\\\{v_{\rm{f}}} = - \sqrt {\frac{{2mgh}}{{\left( {m + \frac{I}{{{r^2}}}} \right)}}} \\\end{array}$

Ans: Part B

The relationship between angular acceleration and linear acceleration is $a = r\alpha$.