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A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30°. 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 θ of the plane is increased, at what value of θ does the cylinder begin to skid, and not roll perfectly?

Mass of the cylinder, m = 10 kg 

Radius of the cylinder, r = 15 cm = 0.15 m 

Co-efficient of kinetic friction, µ= 0.25 

Angle of inclination, θ = 30° 

Moment of inertia of a solid cylinder about its geometric axis, I = mr2
The various forces acting on the cylinder are shown in the following figure, 



The acceleration of the cylinder is given as, 

    a = mg Sinθ / [m + (I/r2) ] 

      = mg Sinθ / [m + {(1/2)mr2/ r2} ] 

      =  g Sin 30° 

      = × 9.8 × 0.5 

      =  3.27 ms-2

(a) Using Newton’s second law of motion,

  Net force, 
fnet = ma 
 
 mg Sin 30° - f = ma 

                     f = mg Sin 30° - ma 

                       = 10 × 9.8 × 0.5 - 10 × 3.27 

       49 - 32.7 = 16.3 N

(b) During rolling, the instantaneous point of contact with the plane comes to rest.

Hence, the work done against frictional force is zero.


(c) For rolling without skid,

Using the relation, 

                    μ = (1/3) tan θ 

              tan θ = 3μ

                       = 3 × 0.25 

∴ θ = tan-1 (0.75)

      = 36.87
°.


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Read each statement below carefully, and state, with reasons, if it is true or false;
(a) During rolling, the force of friction acts in the same direction as the direction of motion of the CM of the body.


a) False

Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.

Tips: -

a) Frictional force acts opposite to the direction of motion of the centre of mass of a body. In the case of rolling, the direction of motion of the centre of mass is backward. Hence, frictional force acts in the forward direction.


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A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 π rad s-1. Which of the two will start to roll earlier? The co-efficient of kinetic friction is μk = 0.2.

Radii of the ring and the disc, r = 10 cm = 0.1 m

Initial angular speed, ω=10 π rad s–1 

Coefficient of kinetic friction, μk = 0.2 

Initial velocity of both the objects, u = 0 

The frictional force causes motion between two objects. 

As per Newton’s second law of motion,

Frictional force, 
f =ma 

                  μkmg= ma 

where, 

 a = Acceleration produced in the objects 

= Mass 

Therefore,

    
a = μkg                                           … (i)

As per the first equation of motion,

Final velocity of the objects can be obtained as,

v = u + at 

   = 0 + μkgt 

   = μkgt                                             … (ii)

The torque applied by the frictional force will act in perpendicularly outward direction and cause reduction in the initial angular speed. 

Torque, τ= – I α 

where, 

α = Angular acceleration 

μkmgr = –I α  



 
α =                                    ...(iii) 

Using the first equation of rotational motion to obtain the final angular speed, 

ω = ω0 + α

   = ω0 + (-μkmgr / I)t                          ...(iv) 

Rolling starts when linear velocity, 

            v = rω 

Therefore, 

  v = r (ω0 μkmgrt / I)                         ...(v) 

Equating equations (ii) and (v), we get

μkgt = r (

      = rω0 -                           ...(vi)

For the ring, 

       I = mr

∴ μkgt = rω0 -
        = rω0 - μkgt 

2μkgt = rω

∴    t =  

       =

       =0.80 s                                       ...(vii)

For the disc:

          I = mr2

∴    μkgt = rω0 - μkmgr2t / mr

            = rω0 - 2μkgt 

    3μkgt = rω

Therefore,

 
t = 

    =       ...(viii) 

Since td > tr, the disc will start rolling before the ring. 
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Read each statement below carefully, and state, with reasons, if it is true or false
The instantaneous acceleration of the point of contact during rolling is zero.

False

This is because when a body is rolling, its instantaneous acceleration is not equal to zero. Instantaneous acceleration will have some value. 
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Read each statement below carefully, and state, with reasons, if it is true or false;
(b) The instantaneous speed of the point of contact during rolling is zero. 

True

Rolling can be considered as the rotation of a body about an axis passing through the point of contact of the body with the ground. Hence, its instantaneous speed is zero. 


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