1. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds.How many revolutions does the wheel make in attaining the speed of 2000 r.p.m.?
a) 333.4
b) 444.4
c) 555.4
d) 666.4
Explanation: Given N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
Angular acceleration
Let α = Angular acceleration in rad/s2.
We know that
ω = ω0 + α.t
or 209.5 = 0 + α × 20
α = 209.5 / 20 = 10.475 rad/s2
We know that the angular distance moved by the wheel during 2000 r.p.m. (i.e. when ω = 209.5 rad/s),
θ = (ω0 + ω)t/2 = ( 0 + 209.5)20/2 = 2095 rad
Since the angular distance moved by the wheel during one revolution is 2π radians, therefore
number of revolutions made by the wheel,
n = θ /2π = 2095/2π = 333.4
2. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the beginning?
a) 288.6 m/s
b) 388.6 m/s
c) 488.6 m/s
d) 188.6 m/s
Explanation: r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s
Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s
3. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the end of the interval ?
a) 235.5 m/s
b) 335.5 m/s
c) 435.5 m/s
d) 535.5 m/s
Explanation: r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s
Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s
Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s
4. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the normal component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?
a) 2.7 m/s2
b) 3.7 m/s2
c) 4.7 m/s2
d) 5.7 m/s2
Explanation: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s
Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s
Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s
Let α = Constant angular acceleration.
We know that ω = ω0+ α.t
157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s2
Radius corresponding to the middle point,
r = 1.5 /2 = 0.75 m
Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s2
5. A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the tangential component of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?
a) 18287 m/s2
b) 18387 m/s2
c) 18487 m/s2
d) 18587 m/s2
Explanation: Given : r = 1.5 m ; N0 = 1200 r.p.m. or ω0 = 2 π × 1200/60 = 125.7 rad/s ;
N = 1500 r.p.m. or ω = 2 π × 1500/60 = 157 rad/s ; t = 5 s
Linear velocity at the beginning
We know that linear velocity at the beginning,
v0 = r . ω0 = 1.5 × 125.7 = 188.6 m/s
Linear velocity at the end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s
Let α = Constant angular acceleration.
We know that ω = ω0+ α.t
157 = 125.7 + α × 5 or α = (157 – 125.7) /5 = 6.26 rad/s2
Radius corresponding to the middle point,
r = 1.5 /2 = 0.75 m
Tangential acceleration = α. r = 6.26 × 0.75 = 4.7 m/s2
Radial acceleration = ω2 . r = (157)2 0.75 = 18487 m/s2
6. The force which acts along the radius of a circle and directed ____________ the centre of the circle is known as centripetal force
a) away from
b) towards
c) at the
d) none of the mentioned
Explanation: Centripetal force acts radially inwards and is essential for circular motion.
7. The unit of mass moment of inertia in S.I. units is
a) m4
b) kgf-m-s2
c) kg-m2
d) N-m
Explanation: Moment of inertia is the distance, from a give reference, where the whole mass of body is assumed to be concentrated to give the same value of I. The unit of mass moment of inertia in S.I. units is kg-m2.
8. Joule is a unit of
a) force
b) work
c) power
d) none of the mentioned
Explanation: In S.I. system of units, the practical unit of work is N-m. It is the work done by a force of 1 newton, when it displaces a body through 1 metre. The work of 1 N-m is known as joule (briefly written as J ) such that 1 N-m = 1 J.
9. The energy possessed by a body, for doing work by virtue of its position, is called
a) potential energ
b) kinetic energy
c) electrical energy
d) chemical energy
Explanation: Potential energy is the energy possessed by a body for doing work, by virtue of its position.
Kinetic energy is the energy possessed by a body, for doing work, by virtue of its mass and velocity of motion.
10. When a body of mass moment of inertia I (about a given axis) is rotated about that axis with an angular velocity, then the kinetic energy of rotation is
a) 0.5 I.ω
b) I.ω
c) 0.5 I.ω2
d) I.ω2
Explanation: When a body of mass moment of inertia I (about a given axis) is rotated about that axis, with an angular velocity ω, then it possesses some kinetic energy. In this case, Kinetic energy of rotation = 1/ 2I.ω2
When a body has both linear and angular motions e.g. in the locomotive driving wheels and wheels of a moving car, then the total kinetic energy of the body is equal to the sum of kinetic energies of translation and rotation.
Total kinetic energy = 1/ 2mv2 +1/ 2I.ω2