- #1

- 42

- 0

## Homework Statement

A uniform rod of mass M and length L can rotate around point P which is at position x from one end of the rod as shown in the figure. The rod is gently placed on a rough horizontal surface that has a friction coefficient μ. and at t=0 starts rotating with angular velocity ω.

The total time of rotation of the rod is T, which depends on x. Let the maximum and minimum time of rotation be T_M and T_m respectively. Find: [tex] \frac{T_m}{T_M}[/tex]

## The Attempt at a Solution

We need to find the torque that the friction force exerts on the rod and from that find the angular deceleration of the rod. We need to consider the part of the rod that is above the point of rotation and the one below it. Let's take a small mass dm with length ds, we have (λ is the linear mass density of the rod) :

[tex] dF_f = μgdm = μgλds[/tex]

[tex] dτ= μgλsds[/tex]

The total torque is the sum of the torques of the two parts:

[tex] \int_{0}^{L-x} μgλs ds + \int_{0}^{-x} μgλs ds = \frac{1}{2}μgλx^2 + \frac{1}{2}μgλ(L-x)^2 = \frac{1}{2}μgλ(L^2 +2(x^2-Lx)) [/tex]

Now:

[tex] α=\frac{τ}{I}= \frac{\frac{1}{2}μgλ(L^2 +2(x^2-Lx))}{\frac{1}{3}ML^2 +Mx^2}= \frac{μgλ(L^2 +2(x^2-Lx))}{\frac{2}{3}ML^2 +2Mx^2} [/tex]

[tex] T= \frac{ω}{α} = \frac{ω(\frac{2}{3}ML^2 +2Mx^2)}{μgλ(L^2 +2(x^2-Lx))} [/tex]

We take the derivative with respect to x and set it equal to 0:

[tex] \frac{d}{dx} \frac{ω(\frac{2}{3}ML^2 +2Mx^2)}{μgλ(L^2 +2(x^2-Lx))} = \frac{4LMω(L^2+Lx-3x^2)}{3gλμ(L^2-2Lx+2x^2)} = \frac{4L^2ω(L^2+Lx-3x^2)}{3gμ(L^2-2Lx+2x^2)} = 0[/tex]

When we solve for x we get one negative root which isn't physical.

[tex] x= \frac{-L+L\sqrt{13}}{-6} = \frac{L(1-\sqrt{13})}{6} [/tex]

Can someone please tell me where's the mistake?

Last edited: