Question

A wheel is free to rotate about its fixed axle. A spring is attached to one of its spokes a distance r from the axle, as shown in Fig.15-52. (a) Assuming that the wheel is a hoop of mass m and radius R, what is the angular frequencyof small oscillations of this system in terms of m, R, r, and the spring constant k? What is $\mathit{\omega }$if (b) r=Rand (c) r=0?

Short answer

(a) The angular frequency of small oscillations of the spring is $\frac{r}{R}\sqrt{\frac{k}{M}}$.

(b) If r = R, the angular frequency is $\sqrt{\frac{k}{M}}$.

(c) If r = 0, the angular frequency is 0.

Step-by-step solution

The given data

• Mass of the wheel is m
• Radius of the wheel is R
• The spring is attached to the spoke at distance r from the axel.

Understanding the concept of oscillations

The spring is connected to the spoke at some distance from the axle of the wheel. As the wheel rotates slightly, in the clockwise direction, the spring gets stretched. This exerts restoring force on the spring which acts like a torque on the wheel. This torque makes the wheel rotate back in the counterclockwise direction. Thus, the system exhibits small oscillations which are simple harmonic in nature.

Formula:

The stretched force on the string in oscillations, (i)

Here, F is force, k is force constant, x is displacement.$F=-kx$

The torque on a body in oscillations,

$\tau =F.r\mathrm{or}\mathrm{I}\mathrm{\alpha }\mathrm{where}\mathrm{\alpha }=\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}$, and I = moment of inertia (ii)

(a) Calculation of angular frequency of small oscillations

When the wheel rotates through small-angle θ, clockwise, the spring gets stretched through a distance,

Hence, the restoring force using equation (i) will be given as:

$F=-kr\theta$

This force acts in a counterclockwise direction on the wheel at a distance r from the axel which is the axis of rotation for the wheel. Thus, it exerts a torque on the wheel, and using equation (ii) is given by,

role="math" localid="1657281161688" $\tau =k{r}^2\theta ...................\left(a\right)$

Here,$\tau$ is the torque, r is the radius of the wheel, k is the force constant and θ is the angular displacement.

The torque on the wheel is also given by$\tau =\mathrm{I}\mathrm{\alpha }$ where I = moment of inertia of the wheel and α is the angular acceleration.

Assuming the wheel as a hoop, we write$\mathrm{I}=M{R}^2.......................\left(b\right)$

And the angular acceleration is given as

$\alpha =\frac{d^2\theta }{d{t}^2}.........................\left(C\right)$

Thus, substituting values from equations (a), (b), and (c) in equation (ii), we get

$M{R}^2\frac{d^2\theta }{d{t}^2}=k{r}^2\theta$

This represents the differential equation of SHM written in the form:

$\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}={\omega }^2\theta$

Thus, comparing the two equations, we get the angular frequency of the small oscillations as:

$\omega =\sqrt{\frac{k{r}^2}{M{R}^2}}...................\left(\mathrm{d}\right)$

Hence, the value of angular frequency is$\frac{r}{R}\sqrt{\frac{k}{M}}$

b) Calculation of angular frequency at r =R

When, r=R the above expression (d) reduces to:

$\omega =\sqrt{\frac{k}{M}}$

Hence, the value of the angular frequency is$\sqrt{\frac{k}{M}}$

c) Calculation of angular frequency when r =0

When r=0 , the above expression (d) becomes,$\omega =0$

Hence, the value of angular frequency is 0.