Advertisements
Advertisements
Question
What are the conditions in which force can not produce torque?
Advertisements
Solution
The forces intersect (or) passing through the axis of rotation cannot produce torque as the perpendicular distance between the forces is 0 i.e. r = 0.
∴ `vectau = vecr xx vecF = 0`.
APPEARS IN
RELATED QUESTIONS
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.
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?
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, ______
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.
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 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.

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 ______.
A particle of mass ‘m’ is moving in time ‘t’ on a trajectory given by
`vecr = 10alphat^2hati + 5beta(t - 5)hatj`
Where α and β are dimensional constants.
The angular momentum of the particle becomes the same as it was for t = 0 at time t = ______ seconds.
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) ______.
