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Angular Momentum of Spinning Top: Calculation and Observation.

Created by @nathanedwards
 at November 1st 2023, 10:08:18 am.
AP_Physics_1-Questions
#conservation
#angular-momentum
#spinning-top

Question:

A spinning top, with a mass of 0.1 kg and a radius of 0.02 m, is initially at rest. A force is applied tangent to the top at a distance of 0.03 m from its center, causing it to start spinning. The force is applied for 0.5 seconds, producing an angular acceleration of 20 rad/s^2. Determine:

a) The initial angular velocity of the top. b) The final angular velocity of the top after 0.5 seconds. c) The change in angular momentum of the top. d) Whether the conservation of angular momentum is observed in this system.

Answer:

a) The initial angular velocity of the top can be calculated using the equation:

ωi=α⋅t \omega_i = \alpha \cdot t

where [ \omega_i ] is the initial angular velocity, [ \alpha ] is the angular acceleration, and [ t ] is the time interval.

Substituting the given values, we get:

ωi=20 rad/s2⋅0.5 s=10 rad/s \omega_i = 20 \, \text{rad/s}^2 \cdot 0.5 \, \text{s} = 10 \, \text{rad/s}

b) The final angular velocity of the top can be calculated using the equation:

ωf=ωi+α⋅t \omega_f = \omega_i + \alpha \cdot t

where [ \omega_f ] is the final angular velocity.

Substituting the values, we get:

ωf=10 rad/s+20 rad/s2⋅0.5 s=20 rad/s \omega_f = 10 \, \text{rad/s} + 20 \, \text{rad/s}^2 \cdot 0.5 \, \text{s} = 20 \, \text{rad/s}

c) The change in angular momentum can be calculated using the equation:

ΔL=I⋅Δω \Delta L = I \cdot \Delta \omega

where [ \Delta L ] is the change in angular momentum, [ I ] is the moment of inertia of the top, and [ \Delta \omega ] is the change in angular velocity.

The moment of inertia of a solid sphere can be calculated using the equation:

I=25⋅m⋅r2 I = \frac{2}{5} \cdot m \cdot r^2

Substituting the given values, we get:

I=25⋅0.1 kg⋅(0.02 m)2=0.00008 kg m2 I = \frac{2}{5} \cdot 0.1 \, \text{kg} \cdot (0.02 \, \text{m})^2 = 0.00008 \, \text{kg m}^2

Substituting the values, we get:

ΔL=(0.00008 kg m2)⋅(20 rad/s−10 rad/s)=0.0008 kg m2⋅s−1 \Delta L = (0.00008 \, \text{kg m}^2) \cdot (20 \, \text{rad/s} - 10 \, \text{rad/s}) = 0.0008 \, \text{kg m}^2 \cdot \text{s}^{-1}

d) To determine whether the conservation of angular momentum is observed in this system, we compare the change in angular momentum to zero. If the change in angular momentum is zero, then angular momentum is conserved.

ΔL=0.0008 kg m2⋅s−1≠0 \Delta L = 0.0008 \, \text{kg m}^2 \cdot \text{s}^{-1} \neq 0

Since the change in angular momentum is not zero, we conclude that the conservation of angular momentum is not observed in this system.

Thus, the answer to part (a) is ωi=10 rad/s \omega_i = 10 \, \text{rad/s} , the answer to part (b) is ωf=20 rad/s \omega_f = 20 \, \text{rad/s} , the answer to part (c) is ΔL=0.0008 kg m2⋅s−1 \Delta L = 0.0008 \, \text{kg m}^2 \cdot \text{s}^{-1} , and the answer to part (d) is No, the conservation of angular momentum is not observed.

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