Physics Homework Question - Cylinder on a frictional conveyor belt

[LoverOfTennis]

Homework Statement

A solid, uniform cylinder with mass M and radius R is initially at rest. It is gently placed on a conveyor belt moving at constant velocity V as shown above. The cylinder's axis is perpendicular to the belt's velocity. The coefficients of static and kinetic friction are nonzero constants. (Ignore other energy losses such as rolling resistance)

Basically a cylinder on its curvy side is placed on a frictional conveyor belt

What is the final speed of the cylinder's center of mass? (in terms of M, R, and v i guess?)

Homework Equations

Torque = r * f
r * f = I(alpha)

Fnet = ma
Ffriction = ma

The Attempt at a Solution

Ffriction = ma
umg = ma
a = ug

rf = i*alpha
rumg = 1/2 mr^2 * (a / r)
ug = 1/2 a

after that i don't know wat to do. i actually i don't really know wat I am doing at all =/

 

[anathema]

Thank you for your question. Let's work through this problem together.

First, we need to consider the forces acting on the cylinder. There are two main forces: the force of gravity (mg) and the force of friction (Ffriction). Since the cylinder is initially at rest, the net force on it is zero. This means that the force of friction must be equal and opposite to the force of gravity.

Next, we need to consider the torque on the cylinder. Since the cylinder is not rotating, the torque must also be zero. This means that the frictional force must act at the center of mass of the cylinder.

Now, let's consider the motion of the cylinder. Since it is on a conveyor belt moving at a constant velocity, the acceleration of the cylinder is also constant. This means that we can use the equation Fnet = ma to solve for the acceleration of the cylinder. Since we already know that Ffriction = mg, we can substitute that into the equation to get:

Fnet = ma
Ffriction = ma
mg = ma
a = g

Now, we can use the equation v = u + at to find the final velocity of the cylinder's center of mass. Since the cylinder starts at rest (u = 0) and the acceleration is g, we can simplify the equation to v = gt. This means that the final velocity of the cylinder's center of mass is directly proportional to the acceleration due to gravity (g), the time it takes to reach that velocity (t), and the direction of the conveyor belt's velocity (which is also the direction of the cylinder's motion).

So, in summary, the final speed of the cylinder's center of mass (v) is equal to the acceleration due to gravity (g) multiplied by the time it takes to reach that velocity (t) and the direction of the conveyor belt's velocity (v). This can be expressed as:

v = gt

I hope this helps. Let me know if you have any further questions or need clarification on any of the steps. Keep up the good work with your studies!

 

[m2003]

As a physicist, it is important to have a clear understanding of the concepts and equations you are using to solve a problem. In this case, you are on the right track by considering the forces acting on the cylinder, including the frictional force and the net force. However, it appears that you are confused about the equations for torque and rotational motion.

First, let's start by considering the forces acting on the cylinder. As you correctly stated, the frictional force is given by Ffriction = umg, where u is the coefficient of friction and mg is the weight of the cylinder. This force is acting in the opposite direction of the motion of the conveyor belt, so we can write the net force as Fnet = Ffriction - mV, where V is the velocity of the conveyor belt. Since the cylinder is initially at rest, the net force is equal to the mass times the acceleration, so we can write:

ma = Ffriction - mV

Substituting in the expression for Ffriction, we get:

ma = umg - mV

Solving for the acceleration, we get:

a = (umg - mV) / m

Now, let's consider the rotational motion of the cylinder. The torque acting on the cylinder is given by T = rf, where r is the radius of the cylinder and f is the frictional force acting at a distance r from the center of mass. This torque is equal to the moment of inertia (I) times the angular acceleration (alpha), so we can write:

rf = I*alpha

Substituting in the expression for f, we get:

rumg = I*alpha

Since the cylinder is rolling without slipping, we can relate the linear acceleration (a) and the angular acceleration (alpha) using the equation a = r*alpha. Substituting this into the equation above, we get:

rumg = I*(a/r)

Solving for the acceleration, we get:

a = (rumg*r) / I

Now, we can use this expression for the acceleration in our equation for the net force:

(umg - mV) / m = (rumg*r) / I

Solving for V, we get:

V = ugr / (umI + mr^2)

This is the final speed of the cylinder's center of mass, in terms of the given variables. It is important to note that the final