What is the angular momentum (in kg*m*m/s) of a 790 gram symmetrical rotating bar which is 2.85 m long and spinning at 245 rpm about the center of the bar?
245 rev/min* 2 pi radians/rev * 1 min/60s
= 25.66 radians/second = omega
I = (1/12) m L^2 (1/12)(.79)(2.85)^2
I = .535 kg m^2
I omega = P = .535*25.66 = 13.7 kg m^2/s
245 rev/min* 2 pi radians/rev * 1 min/60s
= 25.66 radians/second = omega
I = (1/12) m L^2 (1/12)(.79)(2.85)^2
I = .535 kg m^2
I omega = L = .535*25.66 = 13.7 kg m^2/s
To find the angular momentum of a rotating bar, we need to use the formula:
Angular momentum (L) = moment of inertia (I) × angular velocity (ω)
First, let's find the moment of inertia. The moment of inertia for a symmetrical rotating bar can be calculated using the formula:
Moment of inertia (I) = (1/12) × mass (m) × length (L)^2
Given:
Mass (m) = 790 grams = 0.79 kg
Length (L) = 2.85 m
Substituting the values into the moment of inertia formula:
I = (1/12) × 0.79 kg × (2.85 m)^2
Calculating:
I = (1/12) × 0.79 kg × 8.1225 m^2
= 0.0657846 kg·m^2
Next, let's convert the angular velocity from rpm (rotations per minute) to radians per second (rad/s). We can use the conversion factor:
1 rpm = (2π/60) rad/s
Given:
Angular velocity (ω) = 245 rpm
Converting:
ω = 245 rpm × (2π/60) rad/s
≈ 25.769 radians per second (rad/s)
Now we can calculate the angular momentum using the formula:
L = I × ω
Substituting the values:
L = 0.0657846 kg·m^2 × 25.769 rad/s
Calculating:
L ≈ 1.69792 kg·m^2/s
Therefore, the angular momentum of the rotating bar is approximately 1.69792 kg·m^2/s.
To find the angular momentum (L) of a rotating object, we use the formula:
L = I * ω
Where:
L is the angular momentum,
I is the moment of inertia, and
ω is the angular velocity.
To calculate the moment of inertia (I), we need to know the mass (m) and the length (L) of the object. It is given that the mass of the bar is 790 grams and the length of the bar is 2.85 meters.
First, let's convert the mass to kilograms by dividing it by 1000:
m = 790 grams / 1000 = 0.79 kg
Next, we need to calculate the moment of inertia of a symmetrical rotating bar using the formula:
I = (1/3) * m * L^2
Plugging in the values:
I = (1/3) * 0.79 kg * (2.85 m)^2
Calculating this expression, we find:
I ≈ 1.228 kg·m^2
Now, let's convert the given angular velocity from rpm (rotations per minute) to radians per second. There are 2π radians in one rotation, and 60 seconds in one minute:
ω = (245 rpm) * (2π rad/1 rev) * (1 min/60 s)
Simplifying this, we get:
ω ≈ 25.71 rad/s
Finally, substituting the values of I and ω into the angular momentum formula, we can calculate the angular momentum (L):
L = (1.228 kg·m^2) * (25.71 rad/s)
Evaluating this expression, we find:
L ≈ 31.61 kg·m^2/s
Therefore, the angular momentum of the rotating bar is approximately 31.61 kg·m^2/s.