Solution for Question 13
Let us analyze the kinematics and dynamics of the two interacting cylinders to determine the steady-state angular velocity of cylinder Q.
Figure 1: Top-down perspective showing velocity vectors at the point of contact.
1. Analysis of Velocity Vectors
Let the point of contact between the two cylinders be the origin for our kinematic analysis. We define the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ along the $x$, $y$, and $z$ axes respectively.
- Cylinder P: Its axis is fixed along the vertical $y$-direction. It rotates with angular velocity $\vec{\omega}_P = \omega_P \hat{j}$. At the point of contact (on the surface), the tangential velocity $\vec{v}_P$ is horizontal. $$ \vec{v}_P = v_P \hat{i} = (\omega_P r_P) \hat{i} $$
- Cylinder Q: Its axis makes an angle $\theta$ with the axis of P (the vertical). Therefore, the direction of motion of the surface of Q (which is perpendicular to its axis) makes an angle $\theta$ with the horizontal plane (the direction of $\vec{v}_P$). Let $\vec{v}_Q$ be the velocity of Q at the contact point. Its magnitude is $v_Q = \omega_Q r_Q$.
2. Steady State Condition
The problem states that cylinder Q acquires a constant angular velocity. This implies that the net torque acting on Q about its axis of rotation must be zero.
The only force capable of providing torque is the kinetic friction force $\vec{f}$ at the contact point. The torque is given by $\vec{\tau} = \vec{r} \times \vec{f}$, where $\vec{r}$ is the normal vector from the axis to the contact point.
For the component of torque along the rotation axis to be zero, the friction force $\vec{f}$ must be parallel to the axis of rotation of Q.
Since kinetic friction acts opposite to the relative velocity ($\vec{v}_{rel} = \vec{v}_Q – \vec{v}_P$), the relative velocity vector must also be parallel to the axis of Q.
3. Calculation of Angular Velocity
Expanding the dot product condition derived above:
$$ v_Q^2 – \vec{v}_P \cdot \vec{v}_Q = 0 $$ $$ v_Q^2 = v_P v_Q \cos(\alpha) $$Where $\alpha$ is the angle between the velocity vectors $\vec{v}_P$ and $\vec{v}_Q$. As established from the geometry of the axes, this angle is $\theta$.
$$ v_Q = v_P \cos\theta $$Substituting the angular relationships $v_P = \omega_P r_P$ and $v_Q = \omega_Q r_Q$:
$$ \omega_Q r_Q = (\omega_P r_P) \cos\theta $$ $$ \omega_Q = \frac{\omega_P r_P \cos\theta}{r_Q} $$4. Vector Representation
The axis of cylinder Q lies in the $xy$-plane (assuming the standard orientation where normal is $z$) tilted by $\theta$ from the vertical. The unit vector along Q’s axis is $\hat{n} = \sin\theta \hat{i} + \cos\theta \hat{j}$.
Since the rotation is driven by friction from P, and observing the sense of rotation (gearing action reverses direction roughly), the angular velocity vector $\vec{\omega}_Q$ is directed opposite to the general quadrant of $\vec{\omega}_P$. Thus:
$$ \vec{\omega}_Q = – \omega_Q \hat{n} = -\frac{\omega_P r_P \cos\theta}{r_Q} (\sin\theta \hat{i} + \cos\theta \hat{j}) $$This corresponds to option (b).