Question

A small block of mass \(m\) is in contact with the inner wall of a large hollow cylinder. Assume the coefficient of static friction between the object and the wall of the cylinder is \(\mu\). Initially, the cylinder is at rest, and the block is held in place by a peg supporting its weight. The cylinder starts rotating about its center axis, as shown in the figure, with an angular acceleration of \(\alpha\). Determine the minimum time interval after the cylinder begins to rotate before the peg can be removed without the block sliding against the wall.

Short answer

Answer: The minimum time interval is given by \(t=\sqrt{\frac{\mu g}{\alpha}}\), where \(\mu\) is the coefficient of friction between the block and the wall, \(g\) is the acceleration due to gravity, and \(\alpha\) is the angular acceleration of the cylinder.

Step-by-step solution

Analyze the forces acting on the block

There are two main forces acting on the block: gravitational force and friction force. The gravitational force is given by \(mg\), where \(g\) is the acceleration due to gravity, and the friction force is given by \(\mu N\), where \(N\) is the normal force.

Find the centripetal force needed to keep the block in circular motion

The centripetal force required to keep the block in circular motion is given by \(m\omega^2 r\), where \(\omega\) is the angular velocity of the cylinder and \(r\) is the distance from the rotation axis to the block.

Write the equation for centripetal force in terms of angular acceleration and time

We know that \(\omega = \alpha t\), where \(\alpha\) is the angular acceleration, and \(t\) is the time variable, which we want to find. Replace \(\omega\) in the centripetal force equation: Centripetal force = \(m(\alpha t)^2 r\)

Set the friction force equal to the centripetal force needed for circular motion

For the block to not slide against the wall, the friction force must be equal to (or larger than) the centripetal force required for circular motion: \(\mu N = m(\alpha t)^2 r\) However, since the block is pressed against the wall due to centripetal force and there is no vertical motion, the normal force \(N\) is equal to the gravitational force \(mg\). \(\mu mg = m(\alpha t)^2 r\)

Solve for the time variable t

We can rearrange the equation to solve for \(t\): \(t^2 = \frac{\mu mg}{mr\alpha}\) Divide both mass m and radius r: \(t^2 = \frac{\mu g}{\alpha}\) Finally, take the square root of both sides to find the minimum time interval: \(t = \sqrt{\frac{\mu g}{\alpha}}\) So, the minimum time interval after the cylinder begins to rotate before the peg can be removed without the block sliding against the wall is \(t=\sqrt{\frac{\mu g}{\alpha}}\).

Key concepts

Static Friction

When we talk aboutstatic friction, we're referring to the force that keeps an object at rest when a force is applied to it, preventing it from sliding. It's a force that needs to be overcome for the object to start moving. In our exercise, the static friction between the block and the wall of the rotating cylinder is what's keeping the block from sliding down due to gravity.

The formula for the maximum static frictional force is given by \( F_{\text{static friction}} = \mu N \) where \( \mu \) is the coefficient of static friction and \( N \) is the normal force. The normal force is typically the force exerted by a surface to support the weight of an object; in this case, the normal force is the gravitational force acting on the block, \( mg \).Static friction is crucial because it determines how much force can be applied to an object before it starts moving. If you increase the applied force beyond the static friction, the object will transition to kinetic friction and begin slipping.

Centripetal Force

The concept ofcentripetal force is central to understanding circular motion. It's the force that is directed towards the center of a circle and is responsible for keeping an object moving in a curved path. In our context, it's the force required to maintain the block's circular motion along with the inner wall of the cylinder.

The equation for centripetal force is \( F_{\text{centripetal}} = m\omega^2 r \), where \( m \) is the mass of the object, \( \omega \) is the angular velocity, and \( r \) is the radius of the motion path. To keep the block from sliding when the peg is removed, the static friction between the block and the cylinder wall must provide an equal force to the required centripetal force.When the block starts rotating, static friction is what's pulling it towards the center, thus enabling it to follow the circular path. Without sufficient static friction, the block would slip and not follow the cylinder's rotation.

Angular Acceleration

The rate at which the angular velocity of a rotating object changes is known asangular acceleration. It's analogous to linear acceleration but for rotational motion. In our exercise, the cylinder begins at rest and starts rotating with a constant angular acceleration \( \alpha \), meaning that its angular velocity \( \omega \) increases over time.

The relationship between angular acceleration and angular velocity is defined by the equation \( \omega = \alpha t \), where \( t \) represents time. The faster the angular acceleration, the quicker the angular velocity will increase. This relationship is key to determining how quickly the block can safely move in a circular path attached to the cylinder's wall without the aid of the supporting peg.A steady increase in angular velocity will eventually generate enough centripetal force derived from static friction to keep the block adhesively connected to the cylinder without sliding.

Physics Problem Solving

Solvingphysics problems involves a systematic approach to understanding and applying physics concepts to real-world scenarios. In the problem we've been analyzing, a methodical step-by-step process was employed to arrive at a solution.

First, the forces acting on the block were analyzed to establish an equation for the necessary static friction. Next, the problem required the application of the centripetal force formula and its connection to angular acceleration and time. Finally, with the formulation of appropriate relationships and equations, algebra was used to solve for the unknown variable, time, leading to a practical answer.

• Understand the physical situation and identify relevant forces and motion types.
• Apply appropriate physics formulas and principles.
• Create an equation(s) that relates the physical quantities involved.
• Solve for the desired quantity.
• Interpret your result in the context of the physical situation.

Following this logical sequence of steps enables students to approach physics problems with confidence and effectively find solutions.