Advertisements
Advertisements
प्रश्न
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.
उत्तर
a and b
Explanation:
The angular momentum L of a particle with respect to the origin is defined to be L = r × p where r is the position vector of the particle and p is the linear momentum. The direction of L is perpendicular to both d r and p by the right-hand rule.
For particle 1, I1 = r1 × mv, is out of the plane of the paper and perpendicular to r1 and p(mv) Similarly I2 = r2 × m(– v) is into the plane of the paper and perpendicular to r2 and – p.
Hence, total angular momentum
`l = l_1 + l_2 = r_1 xx mv + (- r_2 xx mv)`
`|l| = mvd_1 - mvd_2` as `d_2 > d_1`, total angular momentum will be inward
Hence, I = mv(d2 – d1) ⊗.
APPEARS IN
संबंधित 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?
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 body is in translational equilibrium under the action of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point?
When a body is weighed on an ordinary balance we demand that the arum should be horizontal if the weights on the two pans are equal. Suppose equal weights are put on the two pans, the arm is kept at an angle with the horizontal and released. Is the torque of the two weights about the middle point (point of support) zero? Is the total torque zero? If so, why does the arm rotate and finally become horizontal?
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
A cubical block of mass m and edge a slides down a rough inclined plane of inclination θ with a uniform speed. Find the torque of the normal force acting on the block about its centre.
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, ______
State conservation of angular momentum.
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:
A rod of mass 'm' hinged at one end is free to rotate in a horizontal plane. A small bullet of mass m/4 travelling with speed 'u' hits the rod and attaches to it at its centre. Find the angular speed of rotation of rod just after the bullet hits the rod 3. [take length of the rod as 'l']