# Machine Kinematics Questions and Answers – Acceleration of a Particle along a Circular Path

This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on “Acceleration of a Particle along a Circular Path”.

1. A wheel accelerates uniformly from rest to 2000 r.p.m. in 20 seconds. What is its angular acceleration?

Explanation: Solution. 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
or α = 209.5 / 20 = 10.475 rad/s2

2. 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) 400
b) 300
c) 333.4
d) 200

Explanation: Solution. Given : N0 = 0 or ω = 0 ; N = 2000 r.p.m. or ω = 2π × 2000/60 = 209.5 rad/s ; t = 20s
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
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

3. The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known
a) tangential component
b) normal component
c) parallel component
d) none of the mentioned

Explanation: The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known tangential component.
The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as normal component.

4. The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as
a) tangential component
b) normal component
c) parallel component
d) none of the mentioned

Explanation: The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known tangential component.
The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path, is known as normal component.

5. When a particle moves along a straight path, then the radius of curvature is
a) infinitely small
b) zero
c) infinitely great
d) none of the mentioned

Explanation: When a particle moves along a straight path, then the radius of curvature is infinitely great. This means that v2/r is zero.

6. When a particle moves with a uniform velocity, then dv/dt will be
a) infinitely small
b) zero
c) infinitely great
d) none of the mentioned

Explanation: When a particle moves with a uniform velocity, then dv/dt will be zero. In other words, there will be no tangential acceleration; but the particle will have only normal or radial or centripetal acceleration.

7. 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 of the interval ?
a) 188.6 m/s
b) 235.5 m/s
c) 300 m/s
d) 400 m/s

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

8. 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 end of the interval ?
a) 188.6 m/s
b) 235.5 m/s
c) 300 m/s
d) 400 m/s

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 end of 5 seconds
We also know that linear velocity after 5 seconds,
v5 = r . ω = 1.5 × 157 = 235.5 m/s

Sanfoundry Global Education & Learning Series – Machine Kinematics.
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