For the analysis of rolling motion, we have three equations :

Figure 2: The sphere has mass “m” and radius “r”.

A solid sphere rolls down an incline

(i) Newton’s second law for translation

m g sin θ - f S = m a m g sin θ - f S = m a

(ii) Newton’s second law for rotation

τ = f S r = I α ⇒ f S r = 2 m r 2 α 5 ⇒ f S = 2 m r α 5 τ = f S r = I α ⇒ f S r = 2 m r 2 α 5 ⇒ f S = 2 m r α 5

(iii) Equation of accelerated rolling

a C = a = α r a C = a = α r

Substituting value of “α” in the equation of rotation,

⇒ f S = 2 m a 5 ⇒ f S = 2 m a 5

Putting this value in the equation of translation,

⇒ m g sin θ - 2 m a 5 = m a ⇒ m g sin θ - 2 m a 5 = m a

⇒ a = 5 g sin θ 7 ⇒ a = 5 g sin θ 7

Hence, option (d) is correct.

The solid sphere rotates clockwise to move up along the incline. The component of gravity along the incline acting downward decelerates translation. The friction, therefore, should appear in such a fashion that (i) there is linear acceleration to counter the deceleration due to gravity and (ii) there is angular deceleration so that equation of accelerated rolling is held.

Figure 4: The sphere has mass “m” and radius “r”.

A solid sphere rolls up an incline

This means that friction acts upward. It accelerates translation in the direction of motion.

Hence, options (a) and (c) are correct.

The time taken to reach bottom depends on the translational velocity. In turn, velocity depends on the acceleration of the bodies during motion.

Figure 6: The block slides and solid sphere rolls down the incline.

A block and a solid sphere along identical inclines

In the first case, the block slides on the smooth plane. Hence, there is no friction between surfaces. The force along the incline is component of force due to gravity in the direction of motion. The translational acceleration of the block is :

a B = g sin θ a B = g sin θ

In the second case, static friction less than maximum static friction operates at the contact point. The net force on the sphere, rolling down the incline, is :

F net = m g sin θ - f S F net = m g sin θ - f S

The corresponding translational acceleration of the sphere is :

a S = g sin θ - f S m a S = g sin θ - f S m

Evidently,

a B > a S a B > a S

Therefore, the block reaches the bottom first.

Hence, option (a) is correct.

Rolling requires a minimum threshold value of friction. If the friction is less than this threshold value, then the solid sphere will slide down instead of rolling down. Now, for the analysis of rolling motion, we have three equations :

Figure 8: The solid sphere rolls down the incline.

A solid sphere along an incline

(i) Newton’s second law for translation

m g sin θ - f S = m a m g sin θ - f S = m a

(ii) Newton’s second law for rotation

τ = f S r = I α ⇒ f S r = 2 m r 2 α 5 ⇒ f S = 2 m r α 5 τ = f S r = I α ⇒ f S r = 2 m r 2 α 5 ⇒ f S = 2 m r α 5

(iii) Equation of accelerated rolling

a C = a = α r a C = a = α r

Substituting value of “α” in the equation of rotation,

⇒ f S = 2 m a 5 ⇒ f S = 2 m a 5

Putting this value in the equation of translation,

⇒ m g sin θ - 2 m a 5 = m a ⇒ m g sin θ - 2 m a 5 = m a

⇒ a = 5 g sin θ 7 ⇒ a = 5 g sin θ 7

Thus,

⇒ f S = 2 m a 5 = 2 m g sin θ 7 ⇒ f S = 2 m a 5 = 2 m g sin θ 7

The limiting friction corresponding to a given coefficient of friction is :

F S = μ N = μ m g cos θ F S = μ N = μ m g cos θ

For rolling to take place, the coefficient of friction be such that limiting friction corresponding to it is greater than the static friction required to maintain rolling,

μ m g cos θ > 2 m g sin θ 7 ⇒ μ > 2 g tan θ 7 μ m g cos θ > 2 m g sin θ 7 ⇒ μ > 2 g tan θ 7

Hence, option (a) is correct.

There is no friction between smooth incline and the solid sphere. The rolling sphere will roll on smooth plane at constant velocity. Acceleration will tend to slide the solid sphere on a smooth plane, where there is no friction. Hence, net force on the sphere along the incline should be zero so that the sphere keeps rolling with constant velocity.

Now, the incline itself is an accelerated frame of reference. For force analysis, we consider a pseudo force to covert the accelerated reference system to an equivalent inertia reference system. The pseudo force acts through the center of mass and its magnitude is :

Figure 10: Pseudo force renders accelerated frame into an inertial frame.

A solid sphere along an incline

F Pseudo = m a F Pseudo = m a

Analyzing force for translation along the incline, we have :

m g sin θ - m a cos θ = 0 m g sin θ - m a cos θ = 0

⇒ a = g tan θ ⇒ a = g tan θ

Hence, option (c) is correct.