How Much Force is Needed to Keep a Drilled Cylinder at Rest?

[physicsboy121]

Homework Statement

A uniform cylinder of radius R and mass 8kg has an off-axis hole drilled through it at 2R/5. Its new mass is 6.5kg. The hole and cylinder are parallel with their centers being the same height. What horizontal force, F, must be applied on the top to keep the cylinder at rest?

Homework Equations

I tried writing a torque equation for static equilibrium. However, I need to know the center of mass for the cylinder with the drilled hole, but I do not know it. Is there a better approach to this problem?

The Attempt at a Solution

Torque_net = (6.5)(distance of center of mass from cylinder's center) - (F)(R)=0

 

[haruspex]

Just think it is a solid cylinder plus a thin cylinder of negative mass.

 

[kastelian]

Or you can imagine another hole, drilled symmetrically to the first, with the same mass taken out. The remaining cylinder will have 5 kg at x=0, one hole is actually filled with mass 1.5 kg at 2R/5. Then it shall be easy to calculate new x_com in terms of R.
Also g shall be involved in the torque calculation..

 

[physicsboy121]

I got it, thanks so much!

 

[HalfLostAndSearching]

Center of mass of a solid cylinder with a drilled hole is a bit tricky to determine, but there is a simpler approach to this problem.

First, we can use the principle of conservation of angular momentum to solve for the force F. Since the cylinder is at rest, the net torque acting on it must be zero. This means that the torque due to the force F must be equal in magnitude and opposite in direction to the torque due to the off-axis hole.

Since the cylinder is at rest, its initial and final angular momentum must be equal. We can use the formula for angular momentum, L = Iω, where I is the moment of inertia and ω is the angular velocity. The moment of inertia of a solid cylinder is given by I = 1/2MR^2.

Initial angular momentum = (8)(1/2)(R)^2(0) = 0
Final angular momentum = (6.5)(1/2)(R)^2(ω)

Since the hole is off-axis, the angular velocity ω will be different for the top and bottom halves of the cylinder. However, since they are parallel, their angular velocities will be equal and opposite. This means that the top half of the cylinder will have an angular velocity ω and the bottom half will have an angular velocity -ω.

Thus, we can write:
Final angular momentum = (6.5)(1/2)(R)^2(ω) + (6.5)(1/2)(R)^2(-ω) = 0

Solving for ω, we get ω = 0.

Now, we can use the formula for torque, τ = Fr, where r is the distance from the axis of rotation to the point where the force is applied. In this case, r = R, since the force is applied at the top of the cylinder.

Thus, we can write:
Torque due to F = (F)(R)
Torque due to hole = (6.5)(R)(2R/5)

Setting these two torques equal to each other and solving for F, we get:
(F)(R) = (6.5)(R)(2R/5)
F = (6.5)(2R/5)
F = 2.6R

Therefore, the horizontal force F required to keep the cylinder at rest is