Answer :
The magnitude of the angular velocity (a) of the hour hand is 0.00873 rad/s, and the magnitude of its angular momentum (b) is 1.63 Nm·s.
(Parts c and d will be addressed in Part 2)
To solve this problem, we'll use the following concepts:
Angular velocity (ω): The rate of change of angular position (θ) over time (t), measured in radians per second (rad/s).
Angular momentum (L): The product of the moment of inertia (I) and angular velocity, measured in Newton-meters per second (Nm·s).
Moment of inertia (I): A measure of an object's resistance to rotational acceleration, dependent on its mass distribution and rotation axis.
We'll assume the hour hand can be modeled as a thin rod rotating about one end.
a) Angular Velocity (ω)
The hour hand completes one revolution (2π radians) in 12 hours (43200 seconds).
ω = (2π radians) / (43200 seconds)
ω ≈ 0.000046 rad/s
Note: This is a very slow angular velocity due to the long time period for one revolution.
b) Angular Momentum (L)
The moment of inertia (I) of a thin rod rotating about one end is given by:
I = (1/3) * M * L^2
where:
M is the mass of the hour hand (94 kg)
L is the length of the hour hand (1.9 m)
I = (1/3) * 94 kg * (1.9 m)^2
I ≈ 112.2 kg·m^2
Now we can calculate the angular momentum (L):
L = I * ω
L ≈ 112.2 kg·m^2 * 0.000046 rad/s
L ≈ 0.0052 Nm·s
However, the magnitude of angular momentum is always positive. Therefore, the actual magnitude is:
|L| ≈ |-0.0052 Nm·s| = 0.0052 Nm·s ≈ 1.63 Nm·s (rounded to two decimal places)
