Ch 12: Rotation of a Rigid Body

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 12: Rotation of a Rigid Body

Problem 75

Chapter 12, Problem 75

The sphere of mass M and radius R in FIGURE P12.75 is rigidly attached to a thin rod of radius r that passes through the sphere at distance (1/2)R from the center. A string wrapped around the rod pulls with tension T. Find an expression for the sphere's angular acceleration. The rod's moment of inertia is negligible.

Verified step by step guidance

Step 1: Identify the forces and torques acting on the system. The tension T in the string creates a torque about the axis of rotation. The torque is given by \( \tau = T \cdot r \), where \( r \) is the radius of the rod.

Step 2: Write the rotational form of Newton's second law, \( \tau = I \cdot \alpha \), where \( \tau \) is the net torque, \( I \) is the moment of inertia of the sphere about the axis of rotation, and \( \alpha \) is the angular acceleration.

Step 3: Calculate the moment of inertia \( I \) of the sphere about the axis of rotation. The sphere's center is offset by \( \frac{1}{2}R \) from the axis, so use the parallel axis theorem: \( I = I_{\text{center}} + M \cdot d^2 \), where \( I_{\text{center}} = \frac{2}{5}M R^2 \) is the moment of inertia about the sphere's center and \( d = \frac{1}{2}R \) is the distance from the center to the axis. Thus, \( I = \frac{2}{5}M R^2 + M \left( \frac{1}{2}R \right)^2 \).

Step 4: Substitute the expression for \( I \) into the torque equation \( \tau = I \cdot \alpha \). Replace \( \tau \) with \( T \cdot r \) and solve for \( \alpha \): \( \alpha = \frac{T \cdot r}{I} \).

Step 5: Simplify the expression for \( \alpha \) by substituting the full expression for \( I \): \( \alpha = \frac{T \cdot r}{\frac{2}{5}M R^2 + M \left( \frac{1}{2}R \right)^2} \). This is the angular acceleration of the sphere.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Torque

Torque is a measure of the rotational force applied to an object, calculated as the product of the force and the distance from the pivot point to the line of action of the force. In this scenario, the tension T in the string creates a torque about the center of the sphere, which influences its angular acceleration. The formula for torque (τ) is τ = r × F, where r is the distance from the pivot and F is the force applied.

Moment of Inertia

Moment of inertia is a property of a body that quantifies its resistance to angular acceleration about a given axis. It depends on the mass distribution relative to the axis of rotation. In this problem, while the rod's moment of inertia is negligible, the sphere's moment of inertia (I = (2/5)MR² for a solid sphere) plays a crucial role in determining how the sphere will respond to the applied torque.

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time, typically denoted by α. It is directly related to torque and moment of inertia through Newton's second law for rotation, which states that τ = Iα. In this case, to find the angular acceleration of the sphere, one must relate the net torque produced by the tension in the string to the moment of inertia of the sphere.

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