Question

A circular hoop of mass m, radius r, and infinitesimal thickness rolls without slipping down a ramp inclined at an angle θ with the horizontal. (Intro 1figure)

$MAD_ia_2_v1.jpg$

part a)

What is the acceleration of the center of the hoop?

Express the acceleration in terms of physical constants and all or some of the quantities m,r,and θ.

part b)

What is the minimum coefficient of (static)friction  needed for the hoop to roll without slipping? Note that it is static and not kinetic friction that is relevant here, since the bottom point on the wheel is not moving relative to the ground(this is the meaning of no slipping).

Express the minimum coefficient of friction in terms of all or some of the given quantities m,r,and θ.

Answer 1

Concepts and reason

The concepts used in this question are rotational motion, torque, and static friction.

Firstly, draw a free-body diagram of the forces acting on the circular hoop and use the torque expression to find the equation of motion of the circular hoop.

Finally, balance the forces along the vertical axis and find the coefficient of static friction.

Fundamentals

The torque due to a force is as follows:

$\tau = rf\sin \phi$

Here, f is the force which causes the torque and r is the radial distance from the axis of rotation to the line of force.

The expression of torque can be written as follows:

$\tau = I\alpha$

Here, I is the moment of inertia of the system and $\alpha$ is the angular acceleration of the system.

The angular acceleration is given by the following expression:

$\alpha = \frac{a}{r}$

Here, a is the tangential acceleration and r is the radius.

The force of static friction is given by the following expression:

$f = {\mu _{\rm{s}}}N$

Here, N is the normal force and ${\mu _{\rm{s}}}$ is the coefficient of static friction.

(a)

The figure 1 shows a free-body diagram of a circular hoop rolling down an inclined plane. The weight mg is acting in downward direction and its components are resolved along the x and y axes. A force of friction f is acting along the negative x-axis.

Balance all the forces along the x-axis and rearrange the expression for f.

$\begin{array}{c}\\mg\sin \theta - f = ma\\\\f = mg\sin \theta - ma\\\end{array}$

The torque is acting on the hoop due to the friction force. The torque due to friction force is given as follows:

$\tau = rf\sin \phi$

The angle between the direction of friction and r is 90 degrees. Substitute 90^o for $\phi$ in the above expression.

$\begin{array}{c}\\\tau = rf\sin {90^{\rm{o}}}\\\\ = rf\\\end{array}$

The torque generates an angular acceleration in the circular hoop. The expression of torque can be written as follows:

$\tau = I\alpha$

Substitute rf for $\tau$ in the above expression.

$\begin{array}{c}\\rf = I\alpha \\\\f = \frac{{I\alpha }}{r}\\\end{array}$

The angular acceleration is given by the following expression:

$\alpha = \frac{a}{r}$

Substitute $\frac{a}{r}$ for $\alpha$ and $\frac{{I\alpha }}{r}$ for f in the equation $f = mg\sin \theta - ma$ and solve for a.

$\begin{array}{c}\\\frac{{I\left( {\frac{a}{r}} \right)}}{r} = mg\sin \theta - ma\\\\I\frac{a}{{{r^2}}} + ma = mg\sin \theta \\\\a\left( {1 + \frac{I}{{m{r^2}}}} \right) = g\sin \theta \\\\a = \frac{{g\sin \theta }}{{\left( {1 + \frac{I}{{m{r^2}}}} \right)}}\\\end{array}$

The moment of inertia of a circular hoop is as follows:

$I = m{r^2}$

Substitute $m{r^2}$ for I in the expression $a = \frac{{g\sin \theta }}{{\left( {1 + \frac{I}{{m{r^2}}}} \right)}}$ .

$\begin{array}{c}\\a = \frac{{g\sin \theta }}{{\left( {1 + \frac{{m{r^2}}}{{m{r^2}}}} \right)}}\\\\ = \frac{{g\sin \theta }}{{\left( {1 + 1} \right)}}\\\\ = \frac{{g\sin \theta }}{2}\\\end{array}$

(b)

Refer figure 1, balance all the forces acting along y-axis.

$N = mg\cos \theta$

The force of friction in calculated in part a is as follows:

$f = \frac{{I\alpha }}{r}$

Substitute $\frac{a}{r}$ for $\alpha$ and $m{r^2}$ for I in the above expression.

$\begin{array}{c}\\f = \frac{{m{r^2}\left( {\frac{a}{r}} \right)}}{r}\\\\ = ma\\\end{array}$

The force of static friction is given by the following expression:

$f = {\mu _{\rm{s}}}N$

Substitute ma for f and $mg\cos \theta$ for N in the above expression.

$ma = {\mu _{\rm{s}}}mg\cos \theta$

Substitute $\frac{{g\sin \theta }}{2}$ for a in the above expression.

$\begin{array}{c}\\m\left( {\frac{{g\sin \theta }}{2}} \right) = {\mu _{\rm{s}}}mg\cos \theta \\\\{\mu _{\rm{s}}} = \frac{{\tan \theta }}{2}\\\end{array}$

Ans: Part a

The acceleration is $\frac{{g\sin \theta }}{2}.$