The Setup
A Fixed Axis and a Restless Platform
Imagine a solid cylinder resting peacefully on a horizontal platform. But there is a crucial twist in this setup: the central axis of the cylinder is completely fixed in space. It cannot translate left or right; it is only free to spin about its own axis.
Meanwhile, the platform beneath it is not sitting still. It is oscillating back and forth in the x-direction, executing Simple Harmonic Motion (SHM) described by the equation x=Acos(ωt). The problem also explicitly states that there is no slipping between the cylinder and the platform. This means the friction between them is strong enough to force the cylinder to spin in perfect sync with the platform's back-and-forth motion. Our goal is to find the maximum torque acting on this cylinder.
Kinematics of the Platform
Riding the Harmonic Wave
To understand the forces and torques at play, we first need to understand the acceleration of the platform. Since the platform is driving the rotation of the cylinder, its acceleration is the root cause of the torque.
We start with the position of the platform:
x=Acos(ωt)
To find the velocity, we take the first derivative with respect to time:
v=dtdx=−Aωsin(ωt)
Taking the derivative one more time gives us the acceleration of the platform, ap:
ap=dtdv=−Aω2cos(ωt)
Because the cosine function fluctuates between 1 and −1, the maximum magnitude of this acceleration is simply the amplitude of the acceleration wave:
amax=Aω2
The "No Slipping" Constraint
Bridging Translation and Rotation
Here is where the physics gets beautiful. The "no slipping" condition is the bridge that connects the linear motion of the platform to the rotational motion of the cylinder.
No slipping means that the point on the cylinder that is in direct contact with the platform must have the exact same velocity and acceleration as the platform itself. Let's call this contact point P.
Because the cylinder's central axis is fixed, point P cannot have any translational acceleration of the center of mass. Its entire acceleration comes purely from the rotation of the cylinder. If the cylinder has an angular acceleration α, the tangential acceleration of point P is given by Rα.
Equating the acceleration of the cylinder's bottom point to the platform's acceleration, we get our master constraint equation:
Rα=ap
To find the maximum torque, we need the maximum angular acceleration. Using our constraint equation, we can write:
αmax=Ramax=RAω2
Dynamics of the Cylinder
The Torque Equation
Now we turn to Newton's Second Law for Rotation. The torque τ acting on a rigid body is equal to its moment of inertia I multiplied by its angular acceleration α:
τ=Iα
To find the maximum torque, we simply use the maximum angular acceleration we just derived:
τmax=Iαmax
We know that for a uniform solid cylinder rotating about its central geometric axis, the moment of inertia is:
I=21MR2
The Final Calculation
Bringing It All Together
We are now ready for the final substitution. Let's plug our expressions for I and αmax into the torque equation:
τmax=(21MR2)(RAω2)
Notice how elegantly the math simplifies. One factor of R in the numerator cancels out with the R in the denominator, leaving us with:
τmax=21MRAω2
This is our final answer.
The Way Forward
What About Friction?
It is always a great habit to ask what is actually providing this torque. In this setup, the only horizontal force acting on the cylinder is the static friction f from the platform. Since this friction acts at a distance R from the fixed axis, the torque is τ=fR.
If we wanted to find the maximum friction force required to prevent slipping, we could simply divide our maximum torque by R:
fmax=Rτmax=21MAω2
This tells us exactly how rough the platform needs to be to ensure the cylinder dances perfectly to its harmonic tune!