DPP - Angular Displacement, Velocity and Acceleration
1. The angular speed of a fly-wheel making 120 revolution/minute is
Correct Answer: (c) 4π rad/sec
Angular speed ω = 2π × (revolutions per second)
Revolutions per second = 120/60 = 2 rev/s
ω = 2π × 2 = 4π rad/s
Revolutions per second = 120/60 = 2 rev/s
ω = 2π × 2 = 4π rad/s
2. A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begin, the tangential acceleration is
Correct Answer: (d) \( \frac{1600}{r} \)
Distance covered in 2 revolutions = 2 × 2πr = 4πr
Using v² = u² + 2as
80² = 0² + 2 × a × 4πr
6400 = 8πra
a = \( \frac{6400}{8\pi r} = \frac{800}{\pi r} \)
Using v² = u² + 2as
80² = 0² + 2 × a × 4πr
6400 = 8πra
a = \( \frac{6400}{8\pi r} = \frac{800}{\pi r} \)
3. The moment of inertia of a body about a given axis is 2.4 kg-m². To produce a rotational kinetic energy of 750 J, an angular acceleration of 5 rad/s² must be applied about that axis for
Correct Answer: (b) 5 sec
Rotational KE = \( \frac{1}{2} I \omega^2 \)
750 = \( \frac{1}{2} \times 2.4 \times \omega^2 \)
ω² = \( \frac{750 \times 2}{2.4} = 625 \)
ω = 25 rad/s
Using ω = αt
25 = 5 × t
t = 5 sec
750 = \( \frac{1}{2} \times 2.4 \times \omega^2 \)
ω² = \( \frac{750 \times 2}{2.4} = 625 \)
ω = 25 rad/s
Using ω = αt
25 = 5 × t
t = 5 sec
4. A wheel is rolling along the ground with a speed of 2 m/s. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to
Correct Answer: (c) \( 2\sqrt{2} \) m/s
For a rolling wheel, velocity at any point is the vector sum of translational velocity (v) and tangential velocity (ωr = v).
For points at extremities of horizontal diameter, these velocities are perpendicular.
Resultant velocity = \( \sqrt{v^2 + v^2} = \sqrt{2v^2} = v\sqrt{2} = 2\sqrt{2} \) m/s
For points at extremities of horizontal diameter, these velocities are perpendicular.
Resultant velocity = \( \sqrt{v^2 + v^2} = \sqrt{2v^2} = v\sqrt{2} = 2\sqrt{2} \) m/s
5. The angular velocity of seconds hand of a watch will be
Correct Answer: (b) \( \frac{\pi}{30} \) rad/s
Seconds hand completes one revolution (2π radians) in 60 seconds.
Angular velocity ω = \( \frac{2\pi}{60} = \frac{\pi}{30} \) rad/s
Angular velocity ω = \( \frac{2\pi}{60} = \frac{\pi}{30} \) rad/s
6. The direction of the angular velocity vector is along
Correct Answer: (d) The axis of rotation
The direction of angular velocity vector is given by the right-hand rule:
Curl the fingers of your right hand in the direction of rotation, and your thumb points in the direction of the angular velocity vector, which is along the axis of rotation.
Curl the fingers of your right hand in the direction of rotation, and your thumb points in the direction of the angular velocity vector, which is along the axis of rotation.
7. In rotational motion of a rigid body, all particles move with
Correct Answer: (c) With different linear velocities and same angular velocities
In rotational motion of a rigid body, all particles have the same angular velocity (ω).
However, linear velocity v = ωr depends on the distance from the axis of rotation, so particles at different radii have different linear velocities.
However, linear velocity v = ωr depends on the distance from the axis of rotation, so particles at different radii have different linear velocities.
8. In a bicycle the radius of rear wheel is twice the radius of front wheel. If rF and rr are the radius, vF and vr are speeds of top most points of wheel, then
Correct Answer: (c) vF = vr
For a rolling wheel without slipping, the velocity of the topmost point is twice the velocity of the center of mass.
Since both wheels are attached to the same bicycle, they have the same velocity of center of mass.
Therefore, vF = vr = 2v (where v is the velocity of the bicycle).
Since both wheels are attached to the same bicycle, they have the same velocity of center of mass.
Therefore, vF = vr = 2v (where v is the velocity of the bicycle).
9. A flywheel gains a speed of 540 r.p.m. in 6 sec. Its angular acceleration will be
Correct Answer: (a) 3π rad/sec²
Initial angular velocity ω0 = 0
Final angular velocity ω = 540 rpm = \( \frac{540 \times 2\pi}{60} = 18\pi \) rad/s
Time t = 6 s
Using ω = ω0 + αt
18π = 0 + α × 6
α = \( \frac{18\pi}{6} = 3\pi \) rad/s²
Final angular velocity ω = 540 rpm = \( \frac{540 \times 2\pi}{60} = 18\pi \) rad/s
Time t = 6 s
Using ω = ω0 + αt
18π = 0 + α × 6
α = \( \frac{18\pi}{6} = 3\pi \) rad/s²
10. A point P on the rim of wheel is initially at rest and in contact with the ground. Find the displacement of the point P if the radius of the wheel is 5 m and the wheel rolls forward through half a revolution
Correct Answer: (d) \( 5\sqrt{\pi^2 + 1} \) m
After half revolution, the point moves from bottom to top position.
Horizontal displacement = πr = 5π m
Vertical displacement = 2r = 10 m
Resultant displacement = \( \sqrt{(5\pi)^2 + 10^2} = 5\sqrt{\pi^2 + 4} \) m
Horizontal displacement = πr = 5π m
Vertical displacement = 2r = 10 m
Resultant displacement = \( \sqrt{(5\pi)^2 + 10^2} = 5\sqrt{\pi^2 + 4} \) m
11. A car is moving at a speed of 72 km/hr. The diameter of its wheels is 0.25 m. If the wheels are stopped in 20 rotations by applying brakes, then angular retardation produced by the brakes is
Correct Answer: (a) -25.5 rad/s²
Initial speed v = 72 km/h = 20 m/s
Radius r = 0.25/2 = 0.125 m
Initial angular velocity ω0 = v/r = 20/0.125 = 160 rad/s
Final angular velocity ω = 0
Angular displacement θ = 20 rotations = 20 × 2π = 40π rad
Using ω² = ω0² + 2αθ
0 = (160)² + 2α(40π)
α = - (25600)/(80π) ≈ -25.5 rad/s²
Radius r = 0.25/2 = 0.125 m
Initial angular velocity ω0 = v/r = 20/0.125 = 160 rad/s
Final angular velocity ω = 0
Angular displacement θ = 20 rotations = 20 × 2π = 40π rad
Using ω² = ω0² + 2αθ
0 = (160)² + 2α(40π)
α = - (25600)/(80π) ≈ -25.5 rad/s²
12. The wheel of a car is rotating at the rate of 1200 revolutions per minute. On pressing the accelerator for 10 seconds, it starts rotating at 4500 revolutions per minute. The angular acceleration of the wheel is
Correct Answer: (d) 1980 deg/s²
Initial angular velocity ω0 = 1200 rpm = \( \frac{1200 \times 2\pi}{60} = 40\pi \) rad/s
Final angular velocity ω = 4500 rpm = \( \frac{4500 \times 2\pi}{60} = 150\pi \) rad/s
Time t = 10 s
Using ω = ω0 + αt
150π = 40π + α × 10
α = \( \frac{110\pi}{10} = 11\pi \) rad/s²
Converting to deg/s²: 11π rad/s² = 11π × \( \frac{180}{\pi} \) = 1980 deg/s²
Final angular velocity ω = 4500 rpm = \( \frac{4500 \times 2\pi}{60} = 150\pi \) rad/s
Time t = 10 s
Using ω = ω0 + αt
150π = 40π + α × 10
α = \( \frac{110\pi}{10} = 11\pi \) rad/s²
Converting to deg/s²: 11π rad/s² = 11π × \( \frac{180}{\pi} \) = 1980 deg/s²
13. A torque of 50 Nm acting on a wheel at rest rotates it through 200 radians in 5 sec. Calculate the angular acceleration produced
Correct Answer: (c) 16 rad/sec²
Using θ = ω0t + \( \frac{1}{2} \)αt²
200 = 0 + \( \frac{1}{2} \) × α × (5)²
200 = \( \frac{25}{2} \)α
α = \( \frac{200 \times 2}{25} = 16 \) rad/s²
200 = 0 + \( \frac{1}{2} \) × α × (5)²
200 = \( \frac{25}{2} \)α
α = \( \frac{200 \times 2}{25} = 16 \) rad/s²
14. A constant torque of 1000 N-m turns a wheel of moment of inertia 200 kg-m² about an axis through its centre. Its angular velocity after 3 sec is
Correct Answer: (d) 15 rad/sec
Torque τ = Iα
1000 = 200 × α
α = 5 rad/s²
Using ω = ω0 + αt
ω = 0 + 5 × 3 = 15 rad/s
1000 = 200 × α
α = 5 rad/s²
Using ω = ω0 + αt
ω = 0 + 5 × 3 = 15 rad/s
15. A wheel is rotating at 900 r.p.m. about its axis. When the power is cut-off, it comes to rest in 1 minute. The angular retardation in radian/s² is
Correct Answer: (a) \( \frac{\pi}{2} \)
Initial angular velocity ω0 = 900 rpm = \( \frac{900 \times 2\pi}{60} = 30\pi \) rad/s
Final angular velocity ω = 0
Time t = 1 minute = 60 s
Using ω = ω0 + αt
0 = 30π + α × 60
α = \( -\frac{30\pi}{60} = -\frac{\pi}{2} \) rad/s²
Retardation = \( \frac{\pi}{2} \) rad/s²
Final angular velocity ω = 0
Time t = 1 minute = 60 s
Using ω = ω0 + αt
0 = 30π + α × 60
α = \( -\frac{30\pi}{60} = -\frac{\pi}{2} \) rad/s²
Retardation = \( \frac{\pi}{2} \) rad/s²
16. When a ceiling fan is switched on, it makes 10 revolutions in the first 3 seconds. Assuming a uniform angular acceleration, how many rotation it will make in the next 3 seconds
Correct Answer: (c) 30
Using θ = ω0t + \( \frac{1}{2} \)αt²
For first 3 seconds: 10 × 2π = 0 + \( \frac{1}{2} \) × α × (3)²
20π = \( \frac{9}{2} \)α
α = \( \frac{40\pi}{9} \) rad/s²
For first 6 seconds: θ = 0 + \( \frac{1}{2} \) × \( \frac{40\pi}{9} \) × (6)² = \( \frac{1}{2} \) × \( \frac{40\pi}{9} \) × 36 = 80π rad
Number of revolutions in 6 seconds = \( \frac{80\pi}{2\pi} = 40 \)
Revolutions in next 3 seconds = 40 - 10 = 30
For first 3 seconds: 10 × 2π = 0 + \( \frac{1}{2} \) × α × (3)²
20π = \( \frac{9}{2} \)α
α = \( \frac{40\pi}{9} \) rad/s²
For first 6 seconds: θ = 0 + \( \frac{1}{2} \) × \( \frac{40\pi}{9} \) × (6)² = \( \frac{1}{2} \) × \( \frac{40\pi}{9} \) × 36 = 80π rad
Number of revolutions in 6 seconds = \( \frac{80\pi}{2\pi} = 40 \)
Revolutions in next 3 seconds = 40 - 10 = 30
17. In which case application of angular velocity is useful
Correct Answer: (a) When a body is rotating
Angular velocity is a measure of rotational motion. It is defined as the rate of change of angular displacement and is useful when describing the motion of rotating bodies.
18. When a steady torque is acting on a body, the body
Correct Answer: (c) Gets angular acceleration
According to Newton's second law for rotation, torque τ = Iα, where α is angular acceleration.
A steady torque produces a constant angular acceleration in the body.
A steady torque produces a constant angular acceleration in the body.
19. A wheel has a speed of 1200 revolutions per minute and is made to slow down at a rate of 4 radians/s². The number of revolutions it makes before coming to rest is
Correct Answer: (c) 314
Initial angular velocity ω0 = 1200 rpm = \( \frac{1200 \times 2\pi}{60} = 40\pi \) rad/s
Final angular velocity ω = 0
Angular acceleration α = -4 rad/s²
Using ω² = ω0² + 2αθ
0 = (40π)² + 2 × (-4) × θ
0 = 1600π² - 8θ
θ = \( \frac{1600\pi²}{8} = 200\pi² \) rad
Number of revolutions = \( \frac{200\pi²}{2\pi} = 100\pi ≈ 314 \)
Final angular velocity ω = 0
Angular acceleration α = -4 rad/s²
Using ω² = ω0² + 2αθ
0 = (40π)² + 2 × (-4) × θ
0 = 1600π² - 8θ
θ = \( \frac{1600\pi²}{8} = 200\pi² \) rad
Number of revolutions = \( \frac{200\pi²}{2\pi} = 100\pi ≈ 314 \)
20. A mass is revolving in a circle which is in the plane of paper. The direction of angular acceleration is
Correct Answer: (a) Upward the radius
Angular acceleration is the rate of change of angular velocity. Its direction is perpendicular to the plane of rotation, following the right-hand rule.
For a mass revolving in the plane of paper, angular acceleration is perpendicular to the paper - either upward or downward depending on whether the rotation is speeding up or slowing down.
For a mass revolving in the plane of paper, angular acceleration is perpendicular to the paper - either upward or downward depending on whether the rotation is speeding up or slowing down.
21. A wheel rotates with a constant acceleration of 2.0 radian/sec². If the wheel starts from rest the number of revolutions it makes in the first ten seconds will be approximately
Correct Answer: (b) 16
Using θ = ω0t + \( \frac{1}{2} \)αt²
θ = 0 + \( \frac{1}{2} \) × 2 × (10)² = 100 rad
Number of revolutions = \( \frac{100}{2\pi} ≈ 15.9 ≈ 16 \)
θ = 0 + \( \frac{1}{2} \) × 2 × (10)² = 100 rad
Number of revolutions = \( \frac{100}{2\pi} ≈ 15.9 ≈ 16 \)