Advertisements
Advertisements
प्रश्न
Explain why friction is necessary to make the disc in Figure roll in the direction indicated
(a) Give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins.
(b) What is the force of friction after perfect rolling begins?

Advertisements
उत्तर १
A torque is required to roll the given disc. As per the definition of torque, the rotating force should be tangential to the disc. Since the frictional force at point B is along the tangential force at point A, a frictional force is required for making the disc roll.
(a) Force of friction acts opposite to the direction of velocity at point B. The direction of linear velocity at point B is tangentially leftward. Hence, frictional force will act tangentially rightward. The sense of frictional torque before the start of perfect rolling is perpendicular to the plane of the disc in the outward direction.
(b) Since frictional force acts opposite to the direction of velocity at point B, perfect rolling will begin when the velocity at that point becomes equal to zero. This will make the frictional force acting on the disc zero.
उत्तर २
To roll a disc, we require a torque, which can be provided only by a tangential force. As a force of friction is the only tangential force, in this case, it is necessary.
a)As frictional force at B opposes the velocity of point B, which is to the left, the frictional force must be to the right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards.
(b)As frictional force at B decreases the velocity of the point of contact B with the surface, the perfect rolling begins only when a velocity of point B becomes zero. Also, force of friction would become zero at this stage.
संबंधित प्रश्न
Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.
The torque of the weight of any body about any vertical axis is zero. If it always correct?
A heavy particle of mass m falls freely near the earth's surface. What is the torque acting on this particle about a point 50 cm east to the line of motion? Does this torque produce any angular acceleration in the particle?
If several forces act on a particle, the total torque on the particle may be obtained by first finding the resultant force and then taking torque of this resultant. Prove this. Is this result valid for the forces acting on different particles of a body in such a way that their lines of action intersect at a common point?
A ladder is resting with one end on a vertical wall and the other end on a horizontal floor. If it more likely to slip when a man stands near the bottom or near the top?
Equal torques act on the disc A and B of the previous problem, initially both being at rest. At a later instant, the linear speeds of a point on the rim of A and another point on the rim of B are \[\nu_A\] and \[\nu_B\] respectively. We have
Two discs of the same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of the disc with angular velocities ω1 and ω2. They are brought in to contact face to face coinciding with the axis of rotation. The expression for loss of energy during this process is, ______
The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle θ without slipping and slipping down the incline without rolling is, ______
State conservation of angular momentum.
A particle of mass 5 units is moving with a uniform speed of v = `3sqrt 2` units in the XOY plane along the line y = x + 4. Find the magnitude of angular momentum
A particle of mass m is moving in yz-plane with a uniform velocity v with its trajectory running parallel to + ve y-axis and intersecting z-axis at z = a (Figure). The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is ______.

A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω. A person of mass M is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is ______.
Figure shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant, r1 and r2 are their respective position vectors drawn from point A which is in the plane of the parallel lines. Choose the correct options:

- Angular momentum l1 of particle 1 about A is l1 = mvd1
- Angular momentum l2 of particle 2 about A is l2 = mvr2
- Total angular momentum of the system about A is l = mv(r1 + r2)
- Total angular momentum of the system about A is l = mv (d2 − d1)
⊗ represents a unit vector coming out of the page.
⊗ represents a unit vector going into the page.
The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it?
- The forces may be acting radially from a point on the axis.
- The forces may be acting on the axis of rotation.
- The forces may be acting parallel to the axis of rotation.
- The torque caused by some forces may be equal and opposite to that caused by other forces.
A door is hinged at one end and is free to rotate about a vertical axis (Figure). Does its weight cause any torque about this axis? Give reason for your answer.

Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω2 and ω2 are brought into contact face to face with their axes of rotation coincident.
- Does the law of conservation of angular momentum apply to the situation? why?
- Find the angular speed of the two-disc system.
- Calculate the loss in kinetic energy of the system in the process.
- Account for this loss.
A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is `a/3 R^2` ω. The value of a will be:

