Q#1
Let A be a unit vector along the axis of rotation of a purely rotating body and B be a unit vector along the velocity of a particle P of the body away from the axis. The values of A.B is(a) 1 (b) -1 (c) 0 (d) None of these
Answer: (c)
Explanation: The directions of A and B will be perpendicular to each other. Hence
A.B = |A||B|cos90° = 0
Q#2
A body is uniformly rotating about an axis fixed in an inertial frame of reference. Let A be a unit vector along the axis of rotation and B be the unit vector along the resultant force on a particle P of the body away from the axis. The value of A.B is
(a) 1 (b) -1 (c) 0 (d) None of these
Answer: (c)
Explanation: Since the body is uniformly rotating, the resultant force on the particle P will be in the radial direction. So again the directions of A and B will be perpendicular to each other. Hence A.B = |A|.|B|.Cos90° =0.
Q#3
A particle moves with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin
(a) is zero (b) remains constant
(c) goes on increasing (d) goes on decreasing
Answer: (b)
Explanation: Since the particle moves with a constant velocity along a straight line, the distance of the line from the origin (r) remains constant. The linear momentum of the particle mv is constant. So the angular momentum of the particle about the origin = moment of the linear momentum about origin = mvr = Constant.
Q#4
A body is in pure rotation. The linear speed v of a particle, the distance r of the particle from the axis and the angular velocity ω of the body are related as ω = v/r. Thus
(a) ω ∝ 1/r (b) ω ∝ r
(c) ω = 0 (d) ω is independent of r.
Answer: (d)
Explanation: In fact, v ∝ r and ω is the constant of proportionality. Thus v = ωr and since ω is a constant, so it is independent of r.
Q#5
Figure (10-Q2) shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the common axis. The strings supporting A and B do not slip on the wheels. If x and y be the distances traveled by A and B in the same time interval, then
(a) x = 2y (b) x = y (c) y = 2x (d) none of these
Answer: (c)
Explanation: Both the pulleys have same angular speed hence the speed of the string supporting B will be twice that of A because of the double radius. So it will cover twice distance than A in the same time interval.
Q#6
A body is rotating uniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is
(a) vertical
(b) horizontal and skew with the axis
(c) horizontal and intersecting the axis
(d) none of these.
Answer: (c)
Explanation: Since the body is rotating uniformly, a particle not on the axis moves in a uniform circular motion. This circle will be horizontal with its center at the intersecting point with the axis. So it has a radial acceleration and hence a radial resultant force. Thus this force will be horizontal and intersecting the axis.
Q#7
A body is rotating nonuniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is
(a) vertical
(b) horizontal and skew with the axis
(c) horizontal and intersecting the axis
(d) none of these.
Answer: (b)
Since the body is rotating nonuniformly about a vertical axis, a particle not on the axis will move on a circular path in a horizontal plane. It will have both radial and tangential accelerations and hence radial and tangential forces. The resultant force of these two mutually perpendicular forces will be in the horizontal plane but will not cross the axis.
Q#8
Let F be a force acting on a particle having position vector r. Let Г be the torque of this force about the origin, then
(a) rГ = 0 and FГ = 0 (b) rГ = 0 but FГ ≠ 0
(c) rГ ≠ 0 but FГ ≠ 0 (d) rГ ≠ 0 and FГ ≠ 0
Answer: (a)
Г = r x F and the direction of Г is perpendicular to the plane containing r and F i.e. Г is perpendicular to both r and F. So,
rГ = rГcos90° = 0 and FГ = FГ cos90° = 0
Q#8
Let F be a force acting on a particle having position vector r. Let Г be the torque of this force about the origin, then
(a) rГ = 0 and FГ = 0 (b) rГ = 0 but FГ ≠ 0
(c) rГ ≠ 0 but FГ ≠ 0 (d) rГ ≠ 0 and FГ ≠ 0
Answer: (a)
Г = r x F and the direction of Г is perpendicular to the plane containing r and F i.e. Г is perpendicular to both r and F. So,
rГ = rГcos90° = 0 and FГ = FГ cos90° = 0
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