Question 1

The constant torque required to get a wheel of moment of inertia about its axle of 100 kg m2, starting from rest, to rotate at 5 rev/s after 15 s is:
A) 14.0 Nm
B) 39.6 Nm
C) 209 Nm
D) 3142 Nm

Accepted Answer

The angular acceleration of the wheel can be calculated using the formula:

α = (ωf - ωi) / t

where:
ωi = initial angular velocity = 0
ωf = final angular velocity = 5 rev/s = 31.4 rad/s (conversion: 1 rev/s = 2π rad/s)
t = time taken = 15 s

Hence, α = (31.4 - 0) / 15 = 2.09 rad/s2

The torque required to produce this angular acceleration can be calculated using the formula:

τ = Iα

where:
I = moment of inertia about the axle = 100 kg m2
α = angular acceleration = 2.09 rad/s2

Hence, τ = 100 x 2.09 = 209 Nm

Therefore, the correct answer is B) 209 Nm.

Question 2

If the moment of inertia of a slender rod of length L and mass m about an axis through its centre is mL2/12, then its moment of inertia about an end is:
A) mL2/12
B) mL2/3
C) mL2
D) 13mL2/12

Accepted Answer

According to Parallel Axis Theorem, the moment of inertia of a body about an axis through any point is equal to the moment of inertia about a parallel axis through its center of mass plus the product of its mass and the square of the distance between the two axes. Therefore, the moment of inertia of the slender rod about an end is mL2/12 + m(L/2)2 = mL2/12 + mL2/4 = (mL2/3). Therefore, the correct answer is (B).

Question 3

Decide whether each of these statements is True (T) or False
A flywheel has a moment of inertia of 40 kg m2 about its axis and is at rest when a constant torque of 90 N m is applied. After 15 s:
A) (i) T (ii) T
B) (i) T (ii) F
C) (i) F (ii) T
D) (i) F (ii) F

Accepted Answer

The angular acceleration is:α = T/I = 90 Nm / 40 kg m2 = 2.25 rad/s2The angular velocity after 15 s is:ω = αt = 2.25 rad/s2 * 15 s = 33.75 rad/sThe kinetic energy of the flywheel is:K = 1/2 Iω2 = 1/2 * 40 kg m2 * (33.75 rad/s)2 = 22,500 JTherefore, the statements are both true.

Question 4

The rotor of a motor is accelerated from rest by a torque of 250 N m. When it has rotated through 10 revolutions, the gain in angular kinetic energy is:
A) 2.5kJ
B) 5.0kJ
C) 7.85kJ
D) 15.7kJ

Accepted Answer

The answer of The rotor of a motor is accelerated...

Question 5

A flywheel has a mass of 20 kg and a radius of gyration of 100 mm. Its angular kinetic energy when rotating at 10 rev/s is:
A) 2J
B) 20J
C) 790J
D) 7896J

Accepted Answer

The formula for angular kinetic energy is:
KE = (1/2)Iω^2
where I is the moment of inertia and ω is the angular velocity.
The moment of inertia of a flywheel is given by:
I = mr^2
where m is the mass and r is the radius of gyration.
Substituting the given values, we get:
I = (20 kg)(0.1 m)^2 = 0.2 kg m^2
ω = 2πf = 2π(10 rev/s) = 20π rad/s
KE = (1/2)(0.2 kg m^2)(20π rad/s)^2
KE = 790 J
Therefore, the answer is C, 790J.

Quiz 15: Angular Dynamics

The constant torque required to get a wheel of moment of inertia about its axle of 100 kg m2, starting from rest, to rotate at 5 rev/s after 15 s is:

If the moment of inertia of a slender rod of length L and mass m about an axis through its centre is mL2/12, then its moment of inertia about an end is:

Decide whether each of these statements is True (T) or False A flywheel has a moment of inertia of 40 kg m2 about its axis and is at rest when a constant torque of 90 N m is applied. After 15 s:

The rotor of a motor is accelerated from rest by a torque of 250 N m. When it has rotated through 10 revolutions, the gain in angular kinetic energy is:

A flywheel has a mass of 20 kg and a radius of gyration of 100 mm. Its angular kinetic energy when rotating at 10 rev/s is: