Question 294: Centre of Mass and Rotational Motion

Question

The magnitude of 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 5m, the angle between the force and the position vector is (in radians) :

Accepted Answer

2.5 = 1 $$ 	imes $$ 5 sin $$	heta $$
sin$$	heta $$ = 0.5 = $${1 \over 2}$$
$$	heta $$ = $${\pi  \over 6}$$

Answer

$${\pi  \over 8}$$

Answer

$${\pi  \over 6}$$

Answer

$${\pi  \over 4}$$

Answer

$${\pi  \over 3}$$

Centre of Mass and Rotational Motion

The magnitude of 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 5m, the angle between the force and the position vector is (in radians) :

Choose the correct answer: