System of Particles and Rotational Motion Difficult Questions and Answers

System of Particles and Rotational Motion

If a flywheel makes 120 revolutions/minute, then its angular speed will be

rad/sec

Angular speed, ω =	120 × 2π	= 4π rad/sec
	60

Correct Option: C

rad/sec

Angular speed, ω =	120 × 2π	= 4π rad/sec
	60

Two bodies of mass 1 kg and 3 kg have position vectors î - 2ĵ + ̂k and -3î - 2ĵ + ̂k respectively. The centre of mass of this system has a position vector:

The position vector of the centre of mass of two particle system is given by

^→R =	m₁^{→R₁ + m₂→R₂}
	m₁ + m₂

=	1	= [- 8î - 4ĵ + 4̂k] = - 2î - ĵ + ̂k
	4

Correct Option: A

The position vector of the centre of mass of two particle system is given by

^→R =	m₁^{→R₁ + m₂→R₂}
	m₁ + m₂

=	1	= [- 8î - 4ĵ + 4̂k] = - 2î - ĵ + ̂k
	4

Which of the following statements are correct ?
(A) Centre of mass of a body always coincides with the centre of gravity of the body
(B) Centre of mass of a body is the point at which the total gravitational torque on the body is zero
(C) A couple on a body produce both translational and rotation motion in a body
(D) Mechanical advantage greater than one means that small effort can be used to lift a large load

Centre of mass may or may not coincide with centre of gravity. Net torque of gravitational pull is zero about centre of mass.
τ_g = ∑τ_i = ∑r_i × m_ig = 0
Mechanical advantage , M. A.= Load/Effort
If M.A. > 1 ⇒ Load > Effort

Correct Option: D

A solid homogeneous sphere of mass M and radius R is moving on a rough horizontal surface, partly rolling and partly sliding. During this kind of motion of the sphere

Angular momentum about the point of contact with the surface includes the angular momentum about the centre. Due to friction linear momentum is conserved.

Correct Option: B

Angular momentum about the point of contact with the surface includes the angular momentum about the centre. Due to friction linear momentum is conserved.

A solid cylinder of mass m and radius R rolls down an inclined plane of height h without slipping. The speed of its centre of mass when it reaches the bottom is

K.E. =	1	Iω² +	1	mv²
	2		2

K.E. =	1		1	mr²		ω² +	1	mv²
	2		2				2

=	1	mv² +	1	mv² =	3	mv²
	4		2		4

Now, gain in K.E. = Loss in P.E.

3	mv² = mgh ⇒ v - √		4		gh
4			3

Correct Option: B

K.E. =	1	Iω² +	1	mv²
	2		2

K.E. =	1		1	mr²		ω² +	1	mv²
	2		2				2

=	1	mv² +	1	mv² =	3	mv²
	4		2		4

Now, gain in K.E. = Loss in P.E.

3	mv² = mgh ⇒ v - √		4		gh
4			3