# How to solve this problem without using energy considerations? [closed]

This is a problem from my introductory physics textbook:

A cylinder is released from rest from the top of an incline of inclination $$\theta$$ and length $$l$$. If the cylinder rolls without slipping, what will be its speed when it reaches the bottom.

This is of course easy to solve using the principle of conservation of mechanical energy. The change in potential energy, $$mgl\sin \theta$$, must be equal to the kinetic energy at the bottom, i.e., \begin{align} mgl\sin \theta &=\frac{1}{2}I\omega ^2 +\frac{1}{2}mv^2 \\ &= \frac{1}{4}mv^2+\frac{1}{2}mv^2 = \frac{3}{4}mv^2 \end{align} Solving for v, $$v=\sqrt{4/3gl\sin \theta}$$

My question is, how do we solve this problem without using energy considerations, i.e., while taking a force-based approach?

• Can you put your effort into the problem and specify the step where you are having a problem? Nov 15, 2020 at 13:31
• @Young Kindaichi If it were a block the problem would be easy to solve using basic kinematics, but since the object is rolling I don't quite know where to start. Nov 15, 2020 at 13:46
• You need to be more specific. Questions just asking for a calculation are off topic here. What conceptual issues are you having? Nov 15, 2020 at 14:19
• @BioPhysicist Calculating the torque. Please refer to the comments on the answers below. Nov 15, 2020 at 14:22
• The following may help: physics.stackexchange.com/questions/593380/… and physics.stackexchange.com/questions/593854/…. Nov 18, 2020 at 16:01

You use force and torque relationships to do the evaluation. You can find this approach discussed in the Halliday and Resnick Physics textbooks.

It is important to recognize that for rolling without slipping $$v = r \omega$$ and that allows for the relatively simple evaluation using force and torque provided by Ali; this relationship is not true if the object slips.

For a force to do work the force must act through a distance. For rolling without slipping the force of friction does no work because there is no relative motion of the instantaneous point of contact and the surface. Rolling friction provides a force and a torque but does no work. That is why rolling friction has no effect on the energy balance you provided. As another example, for a fluid moving in a pipe assuming the no-slip condition at the pipe walls, the force of friction from the pipe does no work on the fluid.

If the object slips (slides), the energy approach must account for the work done by friction. For a rigid body, all the friction goes into affecting the kinetic energy since there can be no change in the internal energy of a rigid body. You can find the evaluation for the rigid body sliding case- using energy and force/torque- in some physics mechanics textbooks such as Analytical Mechanics by Fowles.

In reality, the object is not rigid and "heating" effects should be considered.

Using torque on the cylinder you can find the angular speed for any time with the given initial conditions but as you can guess it is quite hard to do because you need to solve the differential equation.

• Can you give me a hint on how to calculate the torque on the cylinder? Nov 15, 2020 at 13:58
• @Polemos cylinder doesn't slip and the force from the friction causes it to rotate. Nov 15, 2020 at 14:10
• $\ddot\theta=\text{constant}$ is a fairly easy differential equation to solve. Nov 15, 2020 at 14:18
• @BioPhysicist I don't know how to solve differential equations. Nov 15, 2020 at 14:24
• @Polemos by that way you can get rid of differantial equation but this way of solving such problem is quite problem dependent not a general method. Nov 15, 2020 at 14:39

The general steps are

1. Make a sketch
2. Draw the forces, torques, coordinate system, etc.
3. Write down Newton's second law for each object (translational/rotational)
4. Derive the equations of motion (position and angle as a function of time)
5. Calculate whatever you need to calculate
• Okay, can you tell me how to calculate the cylinder's angular acceleration? I am confused because all the forces acting on it (graviational, normal) pass through the center and the incline is smooth. Nov 15, 2020 at 13:53
• Hint: The cylinder is not slipping. This is the force that you are missing in your sketch Nov 15, 2020 at 14:03
• Thanks. I got $a=2f/m$, where a is the linear acceleration from the frame of reference of the cylinder and f is the force of friction. We can subtract this value from $g\sin \theta$ to get the acceleration of the cylinder along the incline, but how do we calculate $f$ without knowing the coefficient of friction of the surface? Nov 15, 2020 at 14:20
• Your expression for the linear acceleration is not correct. Hint: The friction force appears in both the linear as in the rotational second law of Newton. So you don't new the coefficient of friction. Nov 15, 2020 at 14:25