The large disk-shaped flywheel illustrated below has a radius R of 0.25 m. It is made to spin by the small wheel that contacts it at its rim. The small wheel applies a constant force of 945 N. Friction in the bearing exerts a retarding torque 21 Nm on the large wheel.
If the magnitude of the angular acceleration of the large wheel is 63 rad/s2, what is its mass?
To find the mass of the large wheel, we need to use a formula that relates angular acceleration (α) to torque (τ) and moment of inertia (I).
The formula is: τ = I * α
Given:
Radius of the large wheel (R) = 0.25 m
Force applied by the small wheel (F) = 945 N
Retarding torque due to friction (τ) = 21 Nm
Angular acceleration (α) = 63 rad/s^2
To find the moment of inertia (I), we can use the formula for the torque exerted by the applied force:
τ = F * R
Substituting the given values:
21 Nm = 945 N * 0.25 m
Solving for the moment of inertia:
I = τ / α
= (945 N * 0.25 m) / 63 rad/s^2
= 3.75 Nm / (rad/s^2)
To convert Nm to kg•m^2 (which is the SI unit for moment of inertia), we divide by the square of the angular acceleration constant:
1 Nm = 1 kg•m^2/s^2
1 rad/s^2 = 1 s^-2
I = (3.75 Nm / (rad/s^2)) / (1 s^-2)^2
= 3.75 Nm / s^2
Therefore, the moment of inertia (I) of the large wheel is 3.75 kg•m^2.
The formula to calculate moment of inertia is: I = (1/2) * m * R^2, where m is the mass and R is the radius of the wheel.
Substituting the values:
3.75 kg•m^2 = (1/2) * m * (0.25 m)^2
Simplifying the equation:
3.75 kg•m^2 = (1/2) * m * 0.0625 m^2
3.75 kg•m^2 = 0.03125 m^2 * m
Dividing by 0.03125 m^2:
3.75 kg = m
The mass of the large wheel is 3.75 kg.
To find the mass of the large wheel, we need to use Newton's second law for rotational motion. The equation is:
τ = Iα
Where:
τ = Torque (Nm)
I = Moment of inertia (kg·m²)
α = Angular acceleration (rad/s²)
We have the retarding torque (τ) as 21 Nm and the angular acceleration (α) as 63 rad/s². We need to find the moment of inertia (I) of the large wheel.
The moment of inertia of a disk-shaped object is given by the equation:
I = (1/2) * m * R²
Where:
m = Mass of the object (kg)
R = Radius of the object (m)
Substituting the given value of the radius (R = 0.25 m) into the equation, we have:
I = (1/2) * m * (0.25)²
I = (1/2) * m * 0.0625
I = 0.03125 * m
Now, we can substitute the values of torque (τ = 21 Nm) and angular acceleration (α = 63 rad/s²) into Newton's second law for rotational motion:
τ = Iα
21 = 0.03125 * m * 63
21 = 1.96875 * m
m = 10.67 kg
Therefore, the mass of the large wheel is approximately 10.67 kg.