Question

A uniform cylinder of radius "R" is spun about its axis to an angular velocity of w and then placed into a corner. the coefficient of friction is "k". How many turns will the cylinder make before coming to a stop?

Answer 1

just use normal equation of uniform accelertated bodies
considering a uniform force of friction giving a uniform retarding acceleartion -a
-a= U(co-eff of friction)*g

take final W=0(omega)
find the displacement....s...here will be angular displacement giving us how many radiand were swept in that time 2pi radians makes 1 turn so...find

Answer 2

This is a good question.

So first, let's look at how much angular momentum, L, the cylinder has initially. The moment of inertia is
\[ I = mr^2 \]
and hence
\[ L = mr^2 \omega \]

Now once the cylinder is placed in the corner there are two torques applied to it. One from the horizontal surface, one from the vertical surface. Call them tau_v and tau_h respectively. We will use the subscripts _v and _h throughout for vertical and horizontal

The horizontal torque is relatively straight-forward. In magnitude it is given by
\[ \tau_h = (distance \ from \ center \ of \ mass)(friction \ force \ horizontally) \]
\[ = rF_{f,h} \hbox{ where $F_{f,h}$ is the horizontal friction} \]
\[ = r(kN_h) \]
\[ = rkmg \]

Now the vertical torque in magnitude is
\[ \tau_v = rF_{f,v} = rkN_v \]
The challenge is to find \( N_v \). Once you have those two torques, they are equal to
\[ \tau_h + \tau_v = I \alpha = I \dot{\omega} \]
and you solve the equation to find when \( \omega(t) = 0 \).

Answer 3

thank you both salini and JamesJ