Question

What is the angular momentum of a 2.5 kg uniform cylindrical grinding wheel of radius 13 cm when rotating at 1800rpm ?
L=______ kg*m^2/s

How much torque is required to stop it in 3.0 ?
[t]=______ m*N

Answer 1

Concepts and reason

The concepts used to solve this problem are angular momentum and torque.

In the first part, use the expression of angular momentum. Substitute the moment of inertia and angular velocity for grinding wheel in the expression and find the angular momentum of grinding wheel.

In the second part, use the expression of torque in terms of rate of change of angular momentum and find the value of torque required to stop the wheel.

Fundamentals

The expression of the magnitude of the rotational angular momentum of the grinding wheel is given as follows:

$L = I\omega$

Here, I is the moment of inertia and $\omega$is the angular speed for one complete rotation.

The torque is defined as the rate of change of angular momentum. Mathematically, it is defined as,

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

(a)

The moment of inertia of the grinding wheel about an axis passing through its center and perpendicular to the plane of the wheel is given as follows:

$I = \frac{1}{2}m{R^2}$

Here, m is the mass of the wheel and R is the radius of the wheel.

Substitute 2.5 kg for m and 13 cm for R in the above expression$I = \frac{1}{2}m{R^2}$.

$\begin{array}{c}\\I = \frac{1}{2}\left( {2.5{\rm{ kg}}} \right){\left( {13{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.0{\rm{ cm}}}}} \right)} \right)^2}\\\\ = 0.021{\rm{ kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The angular frequency of the spinning disk is given as follows:

$\begin{array}{c}\\\omega = 1800{\rm{ rpm}}\\\\{\rm{ = 1800}}\left( {\frac{{2\pi }}{{60}}} \right){\rm{ rad/s}}\\\\{\rm{ = 188}}{\rm{.49 rad/s}}\\\end{array}$

Substitute $0.021{\rm{ kg}} \cdot {{\rm{m}}^2}$for I and $188.49{\rm{ rad/s}}$ for $\omega$in the expression$L = I\omega$.

$\begin{array}{c}\\L = \left( {0.021{\rm{ kg}} \cdot {{\rm{m}}^2}} \right)\left( {188.49{\rm{ rad/s}}} \right)\\\\ = 3.982{\rm{ kg}} \cdot {{\rm{m}}^2}/{\rm{s}}\\\end{array}$

(b)

The torque expression is given as follows:

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

Substitute $3.982{\rm{ kg}} \cdot {{\rm{m}}^2}/{\rm{s}}$for dL and 3.0 sec for dt in expression$\tau = \frac{{dL}}{{dt}}$.

$\begin{array}{c}\\\tau = \frac{{3.982{\rm{ kg}} \cdot {{\rm{m}}^2}/{\rm{s}}}}{{3.0{\rm{ s}}}}\\\\ = 1.33{\rm{ N}} \cdot {\rm{m}}\\\end{array}$

Ans: Part a

The magnitude of the angular momentum of the grinding wheel is ${\rm{3}}{\rm{.982 kg}} \cdot {{\rm{m}}^2}/{\rm{s}}$.

Part b

The torque required to stop the wheel in 3.0 s is $1.33{\rm{ N}} \cdot {\rm{m}}$.