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Rotational Dynamics
Short Answer Questions
- Does the angular momentum of a body moving in a circular path change? Give explanation to your answer. Solution:
The angular momentum of the body moving in circular motion relative to the center of the circular path is always constant. i.e., | L~C | = |~r × p~| = mvrsinθ and since, r and v are perpendicular to each other.
Lc = mvr = mr^2 ω ∵ v = rω But relative to any other point except the center, the angular momentum is changing. Let us consider a point Q as in figure. In this case, the angular momentum is changing because r and θ changes throughout the motion. It can also be dealt in another way Since, r and F(centripetal force) are directed in opposite direction, then, τ = ~r × F~ = 0 relative to C and since, dL dtC = τ = 0. This implies, Lc= constant. But relative to any point (say, Q), LQ 6 = constant since ~r and F~ are oriented in different direction throughout the motion.
- If the ice on the polar caps of the earth melts, how will it affect the duration of the day? Explain. Solution: As the polar ice cap melts, the mass of water will spread out. This will in- crease the moment of inertia of the earth and since no external torque acts,
angular momentum is conserved.
i.e., L = Iω As I increases, the angular velocity(ω) of earth decreases to conserve the angular momentum. Since, ω = (^2) Tπ , the time period of rotation of earth then increases resulting in the increase in length of day.
- A ballet dancer stretches her hands when she wants to come to rest. Why? Solution: Since, no external torque acts on the ballet dancer, the angular momentum is conserved. The angular momentum is,
L = Iω
As the ballet dancer stretches her hand, the moment of inertia (I = mr^2 : r is the distance from the axis of rotation) increases and thus the angular velocity of her should decrease to conserve the angular momentum. Due to this reason, ballet dancer stretches her hands to come to rest.
- Why is it easier to hold down a 10 kg body in your hand at your side than to hold it with your arm extended horizontally?(Do it yourself as an exercise!)
- Experienced cooks can tell whether an egg is raw or hard boiled by rolling it down a slope. How is this possible? What are they looking for? Solution: The hard boiled egg acts just like a rigid body while raw egg contains liquid matter in it. When rotated, the liquid is pushed near the surface (away from the center) due to centrifugal force. Therefore, the moment of iner- tia(depends on the distribution of mass relative to axis of rotation) in raw egg is distributed farther from the axis as compared to boiled egg. So, the moment of inertia of raw egg is more. That is why, the acceleration of raw egg is less , a =
gsinθ K^2 R^2 + 1
Here, I=MK^2
compared to hard boiled egg. Hence, by watching the speed and time of egg to reach the bottom of the inclined plane, experienced cooks can distinguish raw eggs and hard boiled eggs.
- If the earth is struck by meteorites, the earth will slow down slightly. Why? Solution: If the earth is struck by meteorites, it’s mass increases thus increasing the moment of inertia. For the angular momentum, L = Iω to be conserved, ω decreases i.e., the earth will slow down and length of day increases.
- An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4 seconds. Find the angular acceleration and the number of revolutions made by the motor in the 4 sec interval. Solution: Given,
initial revolution per second, f 0 = 50060 = 253 rev/sec final revolution per second, f = 20060 = 103 rev/sec angular acceleration, α=? time, t=4 sec
Now, ω = ω 0 + αt 2 πf = 2πf 0 + α × 4
2 π ×
= 2π ×
2 π
= 4α
4 α = − 10 π
α =
− 10 π 4 ∴ α = − 2. 5 π rad/s^2
The angular displacement is,
θ =
ω^2 − ω 02 2 α
=
(2π × 103 )^2 − (2π × 253 )^2 − 2 × 2. 5 π ∴ θ = 146. 61 rad ∵ θ = 2π × n , where n is the number of revolutions. Then, n =
θ 2 π
= 23. 33 rev
- A constant torque of 500 Nm turns a wheel which has a moment of inertia 20 kgm^2 about it’s centre. Find the angular velocity gained in 2 seconds and the kinetic energy gained. Solution: Given,
torque, τ = 500 Nm moment of inertia, I= 20 kgm^2 angular velocity, ω=? Kinetic energy gained, K.E.=? time, t= 2 s initial angular velocity, ω 0 = 0
Now, τ = Iα
α =
τ I ∴ α = 25 rad/s^2
angular velocity is, ω = ω 0 + αt = 25 × 2 ∴ ω = 50 rad/s
Then, the kinetic energy gained is,
K.E. =
Iω^2
=
× 20 × 502
∴ K.E. = 25000 J
- A computer disk drive is turned on starting from the rest and has constant angular acceleration, a) how long did it take to make the first complete rotation, and b) what is it’s angular acceleration? Given that the disk took 0.750 secs for the drive to make it’s second complete revolution. Solution: For the first revolution, i.e., n=1, θ = 2π × n
2 π = ω 0 t +
αt^2
2 π =
αt^2 ∵ ω 0 = 0
4 π = αt^2
For the second revolution, i.e., n=2,
4 π =
α(t + 0.750)^2
From these two expressions,
αt^2 =
α(t + 0.750)^2
2 t^2 = t^2 + 2 × t × 0 .750 + (0.750)^2 t^2 − 1. 5 t − 0 .5625 = 0
When the frictional torque is applied, ω = 0,
α =
ω − ω 0 t′
=
−^53
rad/s^2
Thus, the frictional torque is,
τ ′^ = Iα
= 60 ×
∴ τ ′^ = 1 Nm
- A disc of moment of inertia 0.1 kgm^2 about its center and radius 0. m is released from rest on a plane inclined at 30◦^ to the horizontal. Calculate the angular velocity after it has rolled 2 m down the plane if its mass is 5 kg. Solution: Given,
moment of inertia, I= 0.1 kgm^2 radius, r= 0.2 m angle of inclination, θ = 30◦ distance travelled by disc, s= 2 m mass of disc, m= 5 kg angular velocity, ω=?
When the disc rolls down a inclined plane,
loss in P.E. = mgh = mgssinθ
Total kinetic energy gained is,
mv^2 +
Iω^2
=
mω^2 r^2 +
Iω^2
=
ω^2 (mr^2 + I)
From these two equations,
loss in P.E. = gain in K.E.
mgssinθ =
ω^2 (mr^2 + I)
5 × 10 × 2 × sin 30 ◦^ =
ω^2 [5 × 0. 22 + 0.1]
50 =
ω^2 × 0. 3
ω^2 =
50 × 2
ω =
∴ ω = 18.25 rad/s
- The atoms in the oxygen molecule O 2 may be considered to be point masses separated by a distance of 1. 2 × 10 −^10 m. The molec- ular speed of an oxygen molecule at s.t.p. is 460 m/s. Given that the rotational kinetic energy of the molecule is two-thirds of its translational kinetic energy, calculate its angular velocity at s.t.p. assuming that molecular rotation takes place about an axis through the center of, and perpendicular to, the line joining the atoms. Solution: Since, the molecular rotation takes place about an axis through the center of line joining the atoms,
r =
d 2
= 0. 6 × 10 −^10 m
molecular speed of oxygen, v= 460m/s angular velocity, ω=?
Given, rotational kinetic energy (Erot) is two-thirds of translational kinetic
