A potter’s wheel of radius 53.13 cm and mass 106.4 kg is freely rotating at 47.1 rev/min. The potter can stop the wheel in 5.65 s by pressing a wet rag against the rim and exerting a radially inward force of 66.2 N.
What is the angular acceleration of the wheel?
The only information of any use in this lengthy paragraph is:
wheel stops from 47.1 rev/min in 5.65 sec
we want radians and seconds
47.1 rev/min * 2 pi rad/rev * 1 min/60 s
= 4.93 radians/second
acceleration = change in velocity/change in time
= 4.93 rad/s / 5.65 s
= .873 radians/second^2
Oh, and the answer is negative (deacceleration)
yea i had the answer but forgot the negative, thanks man! really appreciate it!
To find the angular acceleration of the wheel, we need to use the equation:
torque (τ) = moment of inertia (I) * angular acceleration (α)
In this case, the torque is created by the radially inward force applied by the wet rag. The formula for torque is:
τ = force (F) * radius (r)
First, let's find the torque:
τ = F * r
= 66.2 N * 53.13 cm
Next, we need to convert the radius from centimeters to meters:
r = 53.13 cm * (1 m / 100 cm)
Now we can calculate the torque:
τ = 66.2 N * (53.13 cm * (1 m / 100 cm))
To find the moment of inertia (I) of a wheel, we use the equation:
I = 0.5 * mass (m) * radius squared (r^2)
First, let's convert the mass from kg to grams:
m = 106.4 kg * 1000 g/kg
Then we can calculate the moment of inertia:
I = 0.5 * (106.4 kg * 1000 g/kg) * (53.13 cm * (1 m / 100 cm))^2
Now that we have both the torque (τ) and moment of inertia (I), we can find the angular acceleration (α):
τ = I * α
Rearranging the formula to solve for α:
α = τ / I
Finally, we can substitute the values into the formula to calculate the angular acceleration:
α = (66.2 N * 53.13 cm * (1 m / 100 cm)) / (0.5 * (106.4 kg * 1000 g/kg) * (53.13 cm * (1 m / 100 cm))^2)
Simplifying the equation, we can now calculate the angular acceleration.