Question

A sphere is placed rotating with its centre initially at rest in a corner as shown in figure (a) and (b). Coefficient of friction between all surfaces and the sphere is 1/3. Find the ratio of the frictional force $\frac{f_a}{f_b}$ by ground in situations (a) and (b).

• A. 1
• B. $\frac{9}{10}$
• C. $\frac{10}{9}$
• D. none

Answer 1

D

Answer 2

The problem asks for the ratio of the frictional force by the ground in two situations (a) and (b). In both situations, a sphere is placed in a corner, rotating with its center initially at rest. The coefficient of friction $\mu = 1/3$ is the same for all surfaces.

Let $M$ be the mass of the sphere and $R$ be its radius. Since the center of the sphere is initially at rest, the net force on the sphere in both horizontal and vertical directions must be zero at that instant. Since the sphere is rotating, there is relative motion (slipping) at the contact points, so kinetic friction acts. The magnitude of kinetic friction is $f = \mu N$, where $N$ is the normal force.

Situation (a): Clockwise Rotation

1. Forces and their directions:

   • Gravity ($Mg$) acts downwards.
   • Normal force from the ground ($N_g$) acts upwards.
   • Normal force from the wall ($N_w$) acts horizontally to the right (pushing the sphere away from the wall).
   • The sphere rotates clockwise.
     • The point of contact with the ground moves to the right relative to the ground. So, the frictional force from the ground ($f_g$) acts to the left.
     • The point of contact with the wall moves downwards relative to the wall. So, the frictional force from the wall ($f_w$) acts upwards.

2. Equations of motion (for the center of mass, initially at rest):

   • Vertical equilibrium: $N_g + f_w - Mg = 0 \implies N_g + f_w = Mg$
   • Horizontal equilibrium: $N_w - f_g = 0 \implies N_w = f_g$

3. Frictional forces:

   • $f_g = \mu N_g$
   • $f_w = \mu N_w$

4. Solving for $f_a$ (frictional force from the ground): Substitute the friction equations into the equilibrium equations: $N_g + \mu N_w = Mg$ (1) $N_w = \mu N_g$ (2)

   Substitute (2) into (1): $N_g + \mu (\mu N_g) = Mg$ $N_g (1 + \mu^2) = Mg$ $N_g = \frac{Mg}{1 + \mu^2}$

   The frictional force from the ground in situation (a) is $f_a = f_g = \mu N_g$. $f_a = \frac{\mu Mg}{1 + \mu^2}$

   Given $\mu = 1/3$: $f_a = \frac{(1/3)Mg}{1 + (1/3)^2} = \frac{Mg/3}{1 + 1/9} = \frac{Mg/3}{10/9} = \frac{Mg}{3} \times \frac{9}{10} = \frac{3Mg}{10}$

Situation (b): Counter-clockwise Rotation

1. Forces and their directions:

   • Gravity ($Mg$) acts downwards.
   • Normal force from the ground ($N_g'$) acts upwards.
   • Normal force from the wall ($N_w'$) acts horizontally to the right.
   • The sphere rotates counter-clockwise.
     • The point of contact with the ground moves to the left relative to the ground. So, the frictional force from the ground ($f_g'$) acts to the right.
     • The point of contact with the wall moves upwards relative to the wall. So, the frictional force from the wall ($f_w'$) acts downwards.

2. Equations of motion (for the center of mass, initially at rest):

   • Vertical equilibrium: $N_g' - f_w' - Mg = 0 \implies N_g' - f_w' = Mg$
   • Horizontal equilibrium: $N_w' - f_g' = 0 \implies N_w' = f_g'$

3. Frictional forces:

   • $f_g' = \mu N_g'$
   • $f_w' = \mu N_w'$

4. Solving for $f_b$ (frictional force from the ground): Substitute the friction equations into the equilibrium equations: $N_g' - \mu N_w' = Mg$ (3) $N_w' = \mu N_g'$ (4)

   Substitute (4) into (3): $N_g' - \mu (\mu N_g') = Mg$ $N_g' (1 - \mu^2) = Mg$ $N_g' = \frac{Mg}{1 - \mu^2}$

   The frictional force from the ground in situation (b) is $f_b = f_g' = \mu N_g'$. $f_b = \frac{\mu Mg}{1 - \mu^2}$

   Given $\mu = 1/3$: $f_b = \frac{(1/3)Mg}{1 - (1/3)^2} = \frac{Mg/3}{1 - 1/9} = \frac{Mg/3}{8/9} = \frac{Mg}{3} \times \frac{9}{8} = \frac{3Mg}{8}$

Ratio of Frictional Forces

Finally, find the ratio $\frac{f_a}{f_b}$: $\frac{f_a}{f_b} = \frac{\frac{3Mg}{10}}{\frac{3Mg}{8}} = \frac{3Mg}{10} \times \frac{8}{3Mg} = \frac{8}{10} = \frac{4}{5}$

Comparing this with the given options: (A) 1 (B) $\frac{9}{10}$ (C) $\frac{10}{9}$ (D) none

The calculated ratio is $4/5$, which is not among options (A), (B), or (C).

The final answer is D.