Question

A small block on a frictionless horizontal surface has a mass of 2.50×10−2 . It is attached to a massless cord passing through a hole in the surface. (See the figure below .) The block is originally revolving at a distance of 0.300 from the hole with an angular speed of 1.75 . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 . You may treat the block as a particle.

Is angular momentum conserved?

Find the change in kinetic energy of the block, in J.

How much work was done in pulling the cord? in J.

Answer 1

Concepts and reason

The concepts required to solve these questions are kinetic energy, angular momentum and work done.

Initially, use the expression involving rate of change of angular momentum and torque and then use this to check whether the angular momentum is conserved or not.

For the change in kinetic energy, use the expression of rotational kinetic energy and then take the difference of the initial and final kinetic energy.

The work done in pulling the cord is same as that of the change in kinetic energy of the object.

Fundamentals

The expression involving the rate of change of angular momentum to torque is given as follows:

$\tau = \frac{{dL}}{{dt}}$

Here,$\tau$is the externally applied torque for time t and L is the angular momentum.

The moment of inertia I of an object is given as follows:

$I = m{r^2}$

Here, r is the radial distance of the object from the axis of rotation and m is the mass of the object.

The expression for the rotational kinetic energy ${K_E}$is,

${K_E} = \frac{1}{2}m{r^2}{\omega ^2}$

Here, $\omega$ is the angular velocity of the object.

The angular momentum of the object is given as follows:

$L = m\omega {r^2}$

The work – energy theorem states that the work done by the sum of all the forces acting on a point object equals the change in kinetic energy of that object .

(1)

Consider the motion of an object of mass m moving in circular orbit attached to a massless string.

The expression for the rate of change of angular momentum is given below:

$\tau = \frac{{dL}}{{dt}}$

In moving radially inwards, no external torque is acting on the system. Hence, substitute $0{\rm{ N}} \cdot {\rm{m}}$ for $\tau$ in the above given expression as follows:

$\frac{{dL}}{{dt}} = 0$

It implies that as the object moves radially inward, total angular momentum of the system remain conserved.

(2)

Let the angular momentum of the object changes from ${L_1}$ to ${L_2}$ , as the velocity changes from ${\omega _1}$ to ${\omega _2}$ .

Apply the law of conservation of angular momentum and solve for the angular velocity as follows:

$\begin{array}{c}\\{L_1} = {L_2}\\\\mr_1^2{\omega _1} = mr_2^2{\omega _2}\\\\{\omega _2} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}{\omega _1}\\\end{array}$

Here, ${r_1}$ and ${r_2}$ is the radius vector at two instants.

Substitute , $0.3{\rm{ m}}$ for ${r_1}$, $0.15{\rm{ m}}$ for ${r_2}$ and $1.75{\rm{ rad/s}}$ for ${\omega _1}$as follows :

$\begin{array}{c}\\{\omega _2} = {\left( {\frac{{0.3{\rm{ m}}}}{{0.15{\rm{ m}}}}} \right)^2}\left( {{\rm{1}}{\rm{.75 rad/s}}} \right)\\\\ = 7.0{\rm{ rad/s}}\\\end{array}$

The change in kinetic energy $\Delta {K_E}$ of the object is,

$\begin{array}{c}\\\Delta {K_E} = {K_{fE}} - {K_{iE}}\\\\ = \frac{1}{2}mr_2^2{\omega _2}^2 - \frac{1}{2}mr_1^2{\omega _1}^2\\\end{array}$

Here, ${K_{iE}}$and ${K_{fE}}$ are the initial and final kinetic energy of the block.

Substitute $0.3{\rm{ m}}$ for ${r_1}$, $0.15{\rm{ m}}$ for ${r_2}$ , $1.75{\rm{ rad/s}}$ for ${\omega _1}$ , ${\rm{7}}{\rm{.0 rad/s}}$ for ${\omega _2}$and $2.5 \times {10^{ - 2}}{\rm{ kg}}$ for $m$ as follows :

$\begin{array}{c}\\\Delta {K_E} = \frac{1}{2}\left( {2.5 \times {{10}^{ - 2}}{\rm{ kg}}} \right)\left\{ {{{\left( {0.15{\rm{ m}}} \right)}^2}{{\left( {{\rm{7}}{\rm{.0 rad/s}}} \right)}^2} - {{\left( {0.3{\rm{ m}}} \right)}^2}{{\left( {{\rm{1}}{\rm{.75 rad/s}}} \right)}^2}} \right\}\\\\ = 1.03 \times {10^{ - 2}}{\rm{ J}}\\\end{array}$

(3)

According to work energy theorem, the work done in pulling the cord is given by the change in kinetic energy of the block.

Hence, the work done by pulling cord is $1.03 \times {10^{ - 2}}{\rm{ J}}$.

Ans: Part 1

The angular momentum of the block is conserved.