Solution to Question 7
1. Analyzing the Forces on the Blocks
Let $T_1$ be the tension on the side of mass $M$ and $T_2$ be the tension on the side of mass $m$. Assuming the block $M$ moves downwards with acceleration $a$ (since $M > m$), the equations of motion for the two blocks are:
For block $M$:
$$ Mg – T_1 = Ma \quad \dots(1) $$
For block $m$:
$$ T_2 – mg = ma \quad \dots(2) $$
Rearranging these to express tensions in terms of $a$:
$$ T_1 = M(g-a) $$
$$ T_2 = m(g+a) $$
2. Analyzing the Friction on the Axle
The disc rotates about the fixed cylinder (axle). As the disc is light (mass is not given and does not appear in the final answer), the forces acting on the disc must balance, and the torques must balance the frictional torque.
The total downward load on the axle is the sum of the tensions (neglecting the disc’s weight):
$$ N_{\text{load}} = T_1 + T_2 $$
Due to friction between the inner radius $r$ of the disc and the axle, the reaction force from the axle is not purely vertical. The contact point shifts such that the resultant reaction force makes an angle $\phi$ (angle of friction) with the normal radius, where:
$$ \tan \phi = \mu $$
Physics Derivation: Axle Friction & The Friction Circle
Derivation: Axle Friction & The Friction Circle
When a wheel or disc rotates on a fixed axle, friction occurs at the contact surface between the inner radius of the disc ($r$) and the axle. This friction creates a resistive torque that opposes the rotation. The magnitude of this torque depends on the load, the radius, and the coefficient of friction.
1. The Shift of the Contact Point
Consider the disc rotating clockwise while supporting a downward load $N_{\text{load}}$ (which includes the tensions from the ropes).
- If the axle were frictionless, the disc would hang such that the contact point is at the very top (12 o’clock), and the reaction force would be vertical, passing directly through the center of the axle.
- However, due to friction, the rotating disc surface “climbs” or shifts against the axle. Since the disc rotates clockwise, the contact point $P$ shifts to the right by an angle $\phi$.
- At this shifted point $P$, two forces from the axle act on the disc:
- Normal Force ($N$): Directed radially inward towards the center of the disc (or radially outward from the axle center).
- Friction Force ($f$): Directed tangentially, opposing the motion of the disc’s inner surface relative to the axle. Since the disc rotates clockwise, friction points counter-clockwise (up and left in the diagram).
2. Equilibrium and the Angle of Friction
Since the disc is not accelerating linearly (the center stays fixed), the total reaction force $R_{\text{axle}}$ from the axle must exactly balance the total downward load $N_{\text{load}}$.
$$ \vec{R}_{\text{axle}} = – \vec{N}_{\text{load}} $$
This means the resultant reaction $R_{\text{axle}}$ must be vertical.
This resultant vector is the sum of the Normal vector $\vec{N}$ and the Friction vector $\vec{f}$. By definition of the coefficient of friction $\mu$:
$$ f = \mu N $$
From the vector geometry (see diagram), the angle between the Normal force $N$ and the total Resultant force $R_{\text{axle}}$ is called the angle of friction, denoted by $\phi$.
$$ \tan \phi = \frac{f}{N} = \frac{\mu N}{N} = \mu $$
3. The Friction Circle and Torque Calculation
We established that the resultant reaction force $R_{\text{axle}}$ is vertical and originates from the contact point $P$.
However, the point $P$ is shifted from the vertical axis. We need to find the “lever arm” of this vertical force relative to the center of rotation $O$ to calculate the retarding torque.
Geometry of the Lever Arm:
In the right-angled triangle formed by the center $O$, the contact point $P$, and the vertical line dropping from $P$:
- The hypotenuse is the radius of the inner hole, $r$.
- The angle of shift is $\phi$.
- The perpendicular distance (offset) from the center $O$ to the line of action of the force $R_{\text{axle}}$ is:
$$ \text{Lever Arm} = r \sin \phi $$
This distance $r \sin \phi$ defines the radius of a small imaginary circle called the Friction Circle. The total reaction force is always tangent to this circle.
Final Derivation of Torque
The retarding torque $\tau_f$ is the moment of the reaction force about the center:
$$ \tau_f = \text{Force} \times \text{Lever Arm} $$
Since the magnitude of the vertical reaction force is equal to the load ($R_{\text{axle}} = N_{\text{load}}$):
$$ \tau_f = N_{\text{load}} \times (r \sin \phi) $$
Using the identity $\tan \phi = \mu$, we can also express $\sin \phi$ as:
$$ \sin \phi = \frac{\mu}{\sqrt{1+\mu^2}} $$
Thus giving the form used in the problem:
$$ \tau_f = N_{\text{load}} \frac{\mu r}{\sqrt{1+\mu^2}} $$
This resultant force is tangent to a small imaginary circle of radius $r \sin \phi$, often called the friction circle. The retarding torque $\tau_f$ due to friction is therefore:
$$ \tau_f = N_{\text{load}} \times (r \sin \phi) $$
Using trigonometry, since $\tan \phi = \mu$, we have:
$$ \sin \phi = \frac{\mu}{\sqrt{1+\mu^2}} $$
Substituting the load:
$$ \tau_f = (T_1 + T_2) \frac{\mu r}{\sqrt{1+\mu^2}} $$
3. Torque Equation for the Disc
The driving torque is provided by the difference in tensions acting at the outer radius $R$. For a massless pulley, the net torque is zero (Driving Torque = Resisting Torque).
$$ (T_1 – T_2)R = \tau_f $$
Substituting the expression for $\tau_f$:
$$ (T_1 – T_2)R = (T_1 + T_2) \frac{\mu r}{\sqrt{1+\mu^2}} $$
Let us define the constant $\gamma$ as given in the problem solution:
$$ \gamma = \frac{\mu r}{R\sqrt{1+\mu^2}} $$
Dividing the torque equation by $R$, we get:
$$ T_1 – T_2 = (T_1 + T_2) \gamma $$
Rearranging to group like terms:
$$ T_1(1 – \gamma) = T_2(1 + \gamma) \quad \dots(3) $$
4. Solving for Acceleration
Now, substitute the expressions for $T_1$ and $T_2$ from equations (1) and (2) into equation (3):
$$ M(g-a)(1 – \gamma) = m(g+a)(1 + \gamma) $$
Expand the terms:
$$ M(1 – \gamma)g – M(1 – \gamma)a = m(1 + \gamma)g + m(1 + \gamma)a $$
Group terms containing $a$ on the left and terms containing $g$ on the right:
$$ a \left[ m(1 + \gamma) + M(1 – \gamma) \right] = g \left[ M(1 – \gamma) – m(1 + \gamma) \right] $$
Simplify the bracketed terms:
- Coefficient of $a$: $m + m\gamma + M – M\gamma = (M+m) – \gamma(M-m)$
- Coefficient of $g$: $M – M\gamma – m – m\gamma = (M-m) – \gamma(M+m)$
Finally, solving for $a$:
$$ a = \frac{g \{(M-m) – \gamma(M+m)\}}{\{(M+m) – \gamma(M-m)\}} $$
Where:
$$ \gamma = \frac{\mu r}{R\sqrt{1+\mu^2}} $$
Final Answer:
The acceleration of the load $M$ is:
$$ a = \frac{g \{(M-m) – \gamma(M+m)\}}{\{(M+m) – \gamma(M-m)\}} , \text{ where } \gamma = \frac{\mu r}{R\sqrt{1+\mu^2}} $$
Let $T_1$ be the tension on the side of mass $M$ and $T_2$ be the tension on the side of mass $m$. Assuming the block $M$ moves downwards with acceleration $a$ (since $M > m$), the equations of motion for the two blocks are:
For block $M$:
$$ Mg – T_1 = Ma \quad \dots(1) $$For block $m$:
$$ T_2 – mg = ma \quad \dots(2) $$Rearranging these to express tensions in terms of $a$:
$$ T_1 = M(g-a) $$ $$ T_2 = m(g+a) $$The disc rotates about the fixed cylinder (axle). As the disc is light (mass is not given and does not appear in the final answer), the forces acting on the disc must balance, and the torques must balance the frictional torque.
The total downward load on the axle is the sum of the tensions (neglecting the disc’s weight):
$$ N_{\text{load}} = T_1 + T_2 $$Due to friction between the inner radius $r$ of the disc and the axle, the reaction force from the axle is not purely vertical. The contact point shifts such that the resultant reaction force makes an angle $\phi$ (angle of friction) with the normal radius, where:
$$ \tan \phi = \mu $$
Derivation: Axle Friction & The Friction Circle
When a wheel or disc rotates on a fixed axle, friction occurs at the contact surface between the inner radius of the disc ($r$) and the axle. This friction creates a resistive torque that opposes the rotation. The magnitude of this torque depends on the load, the radius, and the coefficient of friction.
Consider the disc rotating clockwise while supporting a downward load $N_{\text{load}}$ (which includes the tensions from the ropes).
- If the axle were frictionless, the disc would hang such that the contact point is at the very top (12 o’clock), and the reaction force would be vertical, passing directly through the center of the axle.
- However, due to friction, the rotating disc surface “climbs” or shifts against the axle. Since the disc rotates clockwise, the contact point $P$ shifts to the right by an angle $\phi$.
- At this shifted point $P$, two forces from the axle act on the disc:
- Normal Force ($N$): Directed radially inward towards the center of the disc (or radially outward from the axle center).
- Friction Force ($f$): Directed tangentially, opposing the motion of the disc’s inner surface relative to the axle. Since the disc rotates clockwise, friction points counter-clockwise (up and left in the diagram).
Since the disc is not accelerating linearly (the center stays fixed), the total reaction force $R_{\text{axle}}$ from the axle must exactly balance the total downward load $N_{\text{load}}$.
$$ \vec{R}_{\text{axle}} = – \vec{N}_{\text{load}} $$This means the resultant reaction $R_{\text{axle}}$ must be vertical.
This resultant vector is the sum of the Normal vector $\vec{N}$ and the Friction vector $\vec{f}$. By definition of the coefficient of friction $\mu$:
$$ f = \mu N $$From the vector geometry (see diagram), the angle between the Normal force $N$ and the total Resultant force $R_{\text{axle}}$ is called the angle of friction, denoted by $\phi$.
$$ \tan \phi = \frac{f}{N} = \frac{\mu N}{N} = \mu $$We established that the resultant reaction force $R_{\text{axle}}$ is vertical and originates from the contact point $P$.
However, the point $P$ is shifted from the vertical axis. We need to find the “lever arm” of this vertical force relative to the center of rotation $O$ to calculate the retarding torque.
Geometry of the Lever Arm:
In the right-angled triangle formed by the center $O$, the contact point $P$, and the vertical line dropping from $P$:
- The hypotenuse is the radius of the inner hole, $r$.
- The angle of shift is $\phi$.
- The perpendicular distance (offset) from the center $O$ to the line of action of the force $R_{\text{axle}}$ is: $$ \text{Lever Arm} = r \sin \phi $$
This distance $r \sin \phi$ defines the radius of a small imaginary circle called the Friction Circle. The total reaction force is always tangent to this circle.
The retarding torque $\tau_f$ is the moment of the reaction force about the center:
$$ \tau_f = \text{Force} \times \text{Lever Arm} $$Since the magnitude of the vertical reaction force is equal to the load ($R_{\text{axle}} = N_{\text{load}}$):
$$ \tau_f = N_{\text{load}} \times (r \sin \phi) $$Using the identity $\tan \phi = \mu$, we can also express $\sin \phi$ as:
$$ \sin \phi = \frac{\mu}{\sqrt{1+\mu^2}} $$Thus giving the form used in the problem:
$$ \tau_f = N_{\text{load}} \frac{\mu r}{\sqrt{1+\mu^2}} $$Using trigonometry, since $\tan \phi = \mu$, we have:
$$ \sin \phi = \frac{\mu}{\sqrt{1+\mu^2}} $$Substituting the load:
$$ \tau_f = (T_1 + T_2) \frac{\mu r}{\sqrt{1+\mu^2}} $$The driving torque is provided by the difference in tensions acting at the outer radius $R$. For a massless pulley, the net torque is zero (Driving Torque = Resisting Torque).
$$ (T_1 – T_2)R = \tau_f $$Substituting the expression for $\tau_f$:
$$ (T_1 – T_2)R = (T_1 + T_2) \frac{\mu r}{\sqrt{1+\mu^2}} $$Let us define the constant $\gamma$ as given in the problem solution:
$$ \gamma = \frac{\mu r}{R\sqrt{1+\mu^2}} $$Dividing the torque equation by $R$, we get:
$$ T_1 – T_2 = (T_1 + T_2) \gamma $$Rearranging to group like terms:
$$ T_1(1 – \gamma) = T_2(1 + \gamma) \quad \dots(3) $$Now, substitute the expressions for $T_1$ and $T_2$ from equations (1) and (2) into equation (3):
$$ M(g-a)(1 – \gamma) = m(g+a)(1 + \gamma) $$Expand the terms:
$$ M(1 – \gamma)g – M(1 – \gamma)a = m(1 + \gamma)g + m(1 + \gamma)a $$Group terms containing $a$ on the left and terms containing $g$ on the right:
$$ a \left[ m(1 + \gamma) + M(1 – \gamma) \right] = g \left[ M(1 – \gamma) – m(1 + \gamma) \right] $$Simplify the bracketed terms:
- Coefficient of $a$: $m + m\gamma + M – M\gamma = (M+m) – \gamma(M-m)$
- Coefficient of $g$: $M – M\gamma – m – m\gamma = (M-m) – \gamma(M+m)$
Finally, solving for $a$:
$$ a = \frac{g \{(M-m) – \gamma(M+m)\}}{\{(M+m) – \gamma(M-m)\}} $$Where:
$$ \gamma = \frac{\mu r}{R\sqrt{1+\mu^2}} $$The acceleration of the load $M$ is:
$$ a = \frac{g \{(M-m) – \gamma(M+m)\}}{\{(M+m) – \gamma(M-m)\}} , \text{ where } \gamma = \frac{\mu r}{R\sqrt{1+\mu^2}} $$