A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30o. The coefficient of static friction μs = 0.25. (a) How much is the force of friction acting on the cylinder ? (b) What is the work done against friction during rolling ? (c) If the inclination q of the plane is increased, at what value of q does the cylinder begin to skid, and not roll perfectly ?
Given: The mass of the cylinder is 10 kg , radius of cylinder is 15 cm , the angle of inclination is 30 ° and the coefficient of static friction is 0.25 .
a)
The free body diagram of the cylinder is shown as,
Figure (1)
Where, m is the mass of the cylinder, g is the gravitational acceleration, f is the friction force acting on the cylinder, m g sin 30 ° is the weight component along the inclined plane and m g cos 30 ° is the weight component along the direction perpendicular to the inclined plane.
The angular acceleration of cylinder is given as,
a = R α α = a R
The net torque acting on the cylinder is,
τ net = f × R
The net torque also given as,
τ net = I α
Therefore,
f × R = I α f = I α R
By substitute the value of α , we get
f = I ( a R ) R = I a R 2 …… (1)
The equilibrium equation of force along the inclined plane is given as,
F net = m g sin 30 ° − f
By substituting the value of F net and f in the above equation, we get
m a = m g sin 30 ° − I a R 2 m a = m g sin 30 ° − ( 1 2 m R 2 ) a R 2 a = 2 3 g sin 30 °
By substituting the value of a in the equation (1), we get
f = ( 1 2 m R 2 ) ( 2 3 g sin 30 ° ) R 2 = 1 3 m g sin 30 ° …… (2)
By substituting the values in the above equation, we get
f = 1 3 × 10 × 9.81 × sin 30 ° = 16.35 N ≈ 16.4 N
Thus, the force of friction acting on the cylinder is 16.4 N.
b)
During pure rolling motion, the net velocity at the point of contact is zero. Therefore, work done against the friction is zero as the friction force acting at the point of contact is zero at each instant.
c)
The friction force acting on the cylinder is,
f = μ s m g cos θ
From the equation (2), the friction force for the given angle of inclination θ is,
f = 1 3 m g sin θ
By substituting the value of f in the above equation, we get
μ s m g cos θ = 1 3 m g sin θ μ s cos θ = 1 3 sin θ tan θ = 3 μ s θ = tan − 1 ( 3 μ s )
By substituting the value in the above equation, we get
θ = tan − 1 ( 3 × 0.25 ) = 36.87 ° ≈ 37 °
Thus, the angle of inclination at which the cylinder begin to skid is 37 ° .
Given: The mass of the cylinder is
a)
The free body diagram of the cylinder is shown as,
Figure (1)
Where,
The angular acceleration of cylinder is given as,
The net torque acting on the cylinder is,
The net torque also given as,
Therefore,
By substitute the value of
The equilibrium equation of force along the inclined plane is given as,
By substituting the value of
By substituting the value of
By substituting the values in the above equation, we get
Thus, the force of friction acting on the cylinder is 16.4 N.
b)
During pure rolling motion, the net velocity at the point of contact is zero. Therefore, work done against the friction is zero as the friction force acting at the point of contact is zero at each instant.
c)
The friction force acting on the cylinder is,
From the equation (2), the friction force for the given angle of inclination
By substituting the value of
By substituting the value in the above equation, we get
Thus, the angle of inclination at which the cylinder begin to skid is
