What is the angular momentum of a 2.7 kg uniform cylindrical grinding wheel of radius 12 cm when rotating at 1600 rpm? How much torque is required to stop it in 4.4 s?
The angular momenturm is I*w, where
I = (1/2) M R^2
and w is the angular veocity.
The required torque*time to stop rotation is I*w. Therefore
(Torque to stop) = (I*w)/(time)
To calculate the angular momentum of an object, you need two pieces of information: the moment of inertia and the angular velocity. The moment of inertia for a cylindrical object can be calculated using the formula:
I = 0.5 * m * r^2
Where:
I = moment of inertia
m = mass of the object
r = radius of the object
The angular velocity can be converted from rpm to radians per second using the formula:
ω = (2π * n) / 60
Where:
ω = angular velocity in radians per second
n = rotational speed in revolutions per minute (rpm)
Once you have the moment of inertia and angular velocity, you can calculate the angular momentum using the formula:
L = I * ω
Where:
L = angular momentum
I = moment of inertia
ω = angular velocity
Let's calculate the angular momentum of the grinding wheel:
Given:
Mass of the grinding wheel (m) = 2.7 kg
Radius of the grinding wheel (r) = 12 cm = 0.12 m
Rotational speed of the grinding wheel (n) = 1600 rpm
Step 1: Calculate the moment of inertia (I):
I = 0.5 * m * r^2
= 0.5 * 2.7 kg * (0.12 m)^2
= 0.1944 kg·m^2
Step 2: Convert the rotational speed from rpm to radians per second:
ω = (2π * n) / 60
= (2π * 1600) / 60
= 167.55 rad/s
Step 3: Calculate the angular momentum (L):
L = I * ω
= 0.1944 kg·m^2 * 167.55 rad/s
≈ 32.56 kg·m^2/s
So, the angular momentum of the grinding wheel is approximately 32.56 kg·m^2/s.
Now, let's calculate the torque required to stop the grinding wheel in 4.4 seconds.
The torque (τ) required to stop a rotating object in a given time can be calculated using the equation:
τ = L / Δt
Where:
τ = torque
L = angular momentum
Δt = time interval
Given:
Angular momentum (L) = 32.56 kg·m^2/s
Time interval (Δt) = 4.4 s
Step 1: Calculate the torque (τ):
τ = L / Δt
= 32.56 kg·m^2/s / 4.4 s
≈ 7.4 kg·m^2/s^2
So, the torque required to stop the grinding wheel in 4.4 seconds is approximately 7.4 kg·m^2/s^2.