What is the angular momentum of a 1.0kg uniform cylindrical grinding wheel of radius 19cm when rotating at 1700rpm ?How much torque is required to stop it in 4.6s ?
r = 0.19 m
I =.5 m r^2 =.5 (1)(.19)^2 = .0181 kg m^2
1700 rev/min * 1 min/60 s * 2 pi rad/rev
= 178 radians/second = omega
L =I * omega .0181 * 178 = 3.22 kg m^2/s
final omega = initial omega - alpha t
alpha (4.6) = -178
alpha = - 38.7 radians/s^2
Torque = I alpha = -.0181 (38.7)
= - 0.7 kg m^2/s^2 = -0.7 Newton meters
To find the angular momentum of the cylindrical grinding wheel, we first need to calculate its moment of inertia. The moment of inertia of a uniform cylindrical object can be calculated using the formula:
I = (1/2) * m * r^2
Where:
- I is the moment of inertia
- m is the mass of the object
- r is the radius of the object
In this case, the mass of the grinding wheel is given as 1.0 kg, and the radius is given as 19 cm (or 0.19 m). Plugging these values into the formula, we can calculate the moment of inertia:
I = (1/2) * 1.0 kg * (0.19 m)^2
I = 0.0361 kg·m^2
The angular momentum (L) of an object can be calculated using the formula:
L = I * ω
Where:
- L is the angular momentum
- I is the moment of inertia
- ω is the angular velocity
The angular velocity (ω) can be calculated from the given rotational speed (1700 rpm). Since 1 revolution is equal to 2π radians, we can convert rpm to radians per second:
ω = (1700 rpm) * (2π rad/1 min) * (1 min/60 s)
ω = 178.38 rad/s
Now, plugging in the values of I and ω, we can calculate the angular momentum:
L = 0.0361 kg·m^2 * 178.38 rad/s
L ≈ 6.44 kg·m^2/s
Thus, the angular momentum of the grinding wheel is approximately 6.44 kg·m^2/s.
Moving on to the second part of the question, to calculate the torque required to stop the grinding wheel in a given time, we can use the formula:
τ = ΔL / Δt
Where:
- τ is the torque
- ΔL is the change in angular momentum
- Δt is the time
The change in angular momentum (ΔL) can be calculated by subtracting the final angular momentum (0, as the wheel is stopped) from the initial angular momentum (6.44 kg·m^2/s). The time (Δt) is given as 4.6 s. Plugging these values into the formula, we can calculate the torque:
τ = (0 - 6.44 kg·m^2/s) / 4.6 s
τ = -1.4 kg·m^2/s^2
Therefore, the torque required to stop the grinding wheel in 4.6 seconds is approximately -1.4 kg·m^2/s^2, where the negative sign indicates that the torque must exert an opposite direction to the angular momentum.