The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

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 Fig. 7.8. Match the following (most appropriate choice):

A uniform sphere of mass m and radius R is placed on a rough horizontal surface (Fig. 7.9). The sphere is struck horizontally at a height h from the floor. Match the following:

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, loses energy by friction.

A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translational plus rotational) equilibrium because no net external force or torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium? How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane? Will you require external forces to sustain the motion?

A door is hinged at one end and is free to rotate about a vertical axis (Fig. 7.10). Does its weight cause any torque about this axis? Give reason for your answer.