Advertisements
Advertisements
प्रश्न
A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s–1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of the angular momentum of the cylinder about its axis?
Advertisements
उत्तर १
Mass of the cylinder, m = 20 kg
Angular speed, ω = 100 rad s–1
Radius of the cylinder, r = 0.25 m
The moment of inertia of the solid cylinder:
`I = (mr^2)/2`
`= 1/2 xx 20 xx (0.25)^2`
= 0.625 `"kg m"^2`
:.Kinetic energy = `1/2Iomega^2`
`= 1/2 xx 6.25 xx (100)^2 = 3125J`
∴Angular momentum, L = Iω
= 6.25 × 100
= 62.5 Js
उत्तर २
M = 20 kg
Angular speed, w = 100 rad s-1; R = 0.25 m
Moment of inertia of the cylinder about its axis =1/2 MR2 = 1/2 x 20 (0.25)2 kg m2 = 0.625 kg m2
Rotational kinetic energy,
Er = 1/2 Iw2 = 1/2 x 0.625 x (100)2 J = 3125 J
Angular momentum, L = Iw = 0.625 x 100 Js= 62.5 Js
संबंधित प्रश्न
Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and 'p_z`. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
The torque of the weight of any body about any vertical axis is zero. If it always correct?
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
The density of a rod gradually decreases from one end to the other. It is pivoted at an end so that it can move about a vertical axis though the pivot. A horizontal force F is applied on the free end in a direction perpendicular to the rod. The quantities, that do not depend on which end of the rod is pivoted, are ________________ .
A simple pendulum of length l is pulled aside to make an angle θ with the vertical. Find the magnitude of the torque of the weight ω of the bob about the point of suspension. When is the torque zero?
When a force of 6⋅0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut?
Calculate the total torque acting on the body shown in the following figure about the point O.
What are the conditions in which force can not produce torque?
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 ______.
Choose the correct alternatives:
- For a general rotational motion, angular momentum L and angular velocity ω need not be parallel.
- For a rotational motion about a fixed axis, angular momentum L and angular velocity ω are always parallel.
- For a general translational motion , momentum p and velocity v are always parallel.
- For a general translational motion, acceleration a and velocity v are always parallel.
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 uniform sphere of mass m and radius R is placed on a rough horizontal surface (Figure). The sphere is struck horizontally at a height h from the floor. Match the following:
| Column I | Column II | |
| (a) h = R/2 | (i) | Sphere rolls without slipping with a constant velocity and no loss of energy. |
| (b) h = R | (ii) | Sphere spins clockwise, loses energy by friction. |
| (c) h = 3R/2 | (iii) | Sphere spins anti-clockwise, loses energy by friction. |
| (d) h = 7R/5 | (iv) | Sphere has only a translational motion, looses energy by friction. |
Angular momentum of a single particle moving with constant speed along the circular path ______.
