Advertisements
Advertisements
Question
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.

Advertisements
Solution
According to the diagram, where weight of the door acts along the negative y-axis.
Torque is not produced by weight about the y-axis.
Because the direction of weight is parallel to the y-axis (axis of rotation).
A force can produce torque only along the direction normal to itself because f = r × F. So when the door is in the xy-plane, the torque produced by gravity can only be along ±z-direction never about an axis passing through the y-direction.
Hence, the weight will not produce any torque about the y-axis.
APPEARS IN
RELATED QUESTIONS
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.
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?

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 rectangular brick is kept on a table with a part of its length projecting out. It remains at rest if the length projected is slightly less than half the total length but it falls down if the length projected is slightly more than half the total length. Give reason.
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?
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.

A 6⋅5 m long ladder rests against a vertical wall reaching a height of 6⋅0 m. A 60 kg man stands half way up the ladder.
- Find the torque of the force exerted by the man on the ladder about the upper end of the ladder.
- Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to be smooth, find the force exerted by the ground on the ladder.
What are the conditions in which force can not produce torque?
State conservation 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 ______.

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.
A uniform cube of mass m and side a is placed on a frictionless horizontal surface. A vertical force F is applied to the edge as shown in figure. Match the following (most appropriate choice):

| (a) mg/4 < F < mg/2 | (i) Cube will move up. |
| (b) F > mg/2 | (ii) Cube will not exhibit motion. |
| (c) F > mg | (iii) Cube will begin to rotate and slip at A. |
| (d) F = mg/4 | (iv) Normal reaction effectively at a/3 from A, no motion. |
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:

The position vector of 1 kg object is `vecr = (3hati - hatj)` m and its velocity `vecv = (3hati + hatk)` ms−1. The magnitude of its angular momentum is `sqrtx` Nm where x is ______.
The magnitude of the torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians) ______.
