Question

A \(95.2-\mathrm{kg}\) solid sphere with a \(14.8-\mathrm{cm}\) radius is suspended by a vertical wire attached to the ceiling of a room. A torque of \(0.192 \mathrm{~N} \cdot \mathrm{m}\) is required to swist the sphere through an angle of \(0.850 \mathrm{rad} .\) Find the period of oscillation when the sphere is released from this position.

Short answer

The period of oscillation of the sphere is calculated using the equation \(T = \sqrt {\theta I/0.192}\) . Substituting the known values into this equation will provide the correct answer.

Step-by-step solution

Calculate the moment of inertia

The moment of inertia \(I\) for a solid sphere is found using the formula \(I = \frac{2}{5}mr^2\) where \(m = 95.2\,kg\) is the mass of the sphere and \(r = 14.8\,cm = 0.148 \,m\) is the radius of the sphere.

Calculate the restoring torque

The restoring torque \(\tau\) can be calculated using the formula \(\tau = I\alpha \) where \(\alpha\) is the angular acceleration. We can rearrange this formula to find \(\alpha = \tau/I = 0.192 \, N \cdot m/I\).

Use Small Angle Approximation

We can use the small angle approximation \(\alpha = \theta / T^2\) where \(T\) is the period of oscillation we want to find and \(\theta = 0.850\, rad\) is the angle. By substituting for \(\alpha\) in Step 2, we obtain \(\theta / T^2 = 0.192 \, N \cdot m/I \) .

Solve for the period \(T\)

Rearrange the last equation to solve for \(T\). From this, we obtain \(T = \sqrt {\theta I/0.192}\) . By substituting the calculated value for \(I\) from Step 1 and \(\theta = 0.850\, rad\), we should get the period of oscillation.

Key concepts

Moment of Inertia

Exploring the moment of inertia, we can imagine it as the rotational equivalent of mass for linear motion. It quantifies an object's resistance to changes in its rotational motion. For a solid sphere, like the one mentioned in the exercise, the moment of inertia, denoted as I, is given by the formula \(I = \frac{2}{5}mr^2\), with m representing the mass, and r representing the radius of the sphere.

When calculating I, it's crucial to ensure that all units are consistent, typically in kilograms for mass and meters for radius or distance. This consistency ensures accuracy when applying the formula to determine the moment of inertia, and therefore, to predict the sphere's rotational behavior.

Restoring Torque

Restoring torque is the force that tends to bring a rotating object back to its equilibrium position. The torque required to twist our sphere by an angle is indicative of how much restoring torque the suspended sphere will experience. In the context of the exercise, the restoring torque \(\tau\) can be expressed through the formula \(\tau = I\alpha\), where \(\alpha\) is the angular acceleration.

Understanding this concept aids in visualizing how the sphere reacts when released: the internal forces within the wire work to restore the sphere to its original, untwisted state, similar to how a spring pulls an object back after being stretched or compressed.

Small Angle Approximation

The small angle approximation is a simplification used in physics to make calculations more manageable while still maintaining accuracy for small angles. According to this approximation, for angles measured in radians, the sine and tangent of the angle can be approximated to the angle itself when the angle is close to zero. That means \(\sin(\theta) \approx \theta\) and \(\tan(\theta) \approx \theta\) when \(\theta\) is small.

In our scenario, the small angle approximation \(\alpha = \theta / T^2\) allows us to link angular acceleration with the period of oscillation T and the angle \(\theta\). This assists in formulating a relationship that is solvable for T, making it possible to predict how long it takes for one complete oscillation under the influence of the restoring torque.

Angular Acceleration

Angular acceleration, denoted by \(\alpha\), measures how quickly the angular velocity of an object changes with time. It plays a pivotal role in the dynamics of rotational motion and is affected by factors such as the applied torque and the moment of inertia. In the context of our oscillating sphere, angular acceleration describes how swiftly the sphere speeds up or slows down in its rotational motion after being released.

By knowing that the restoring torque causes an angular acceleration opposite to the direction of displacement, we can calculate \(\alpha = \tau/I\) when we know the applied torque \(\tau\) and the sphere's moment of inertia \(I\). The angular acceleration is an essential variable in determining the period of oscillation using the small angle approximation, as it informs us about the pace at which the sphere will return to its initial position.