How Does Angular Motion Affect the Centrifugal Force on a Motorcar Wheel?

[Renni202]

A mass of 5g is required to balance a motorcar wheel. If the mass is attached to the 0.33m diameter wheel rim, calculate the centrifugal force acting on the rim of the wheel. The speed of rotation is 521 rev/min.ω= 2∏N/60
F= M ω² T
ω=2∏*521/60 = 54.6 rad/s
F= 5*10^-3*(54.6)²*T

I am not sure how to find T and wonder if anyone can help?

 

[Simon Bridge]

What is T supposed to be?

 

[Renni202]

torque? I am confusing myself thought I thought I would find T by using T=Ia although I would need to find I first.
So am I way out? Is it just half the diameter? 0.165

 

[Simon Bridge]

Which torque?
You are told that the wheel is spinning at a constant speed, and that it is balanced.
What does that tell you about the net torque?

 

[Nickie]

To find T, we need to know the angular acceleration of the wheel. This can be calculated using the formula:

α = ω/t

Where α is the angular acceleration, ω is the angular velocity (which we have calculated to be 54.6 rad/s), and t is the time taken for one revolution (which can be calculated by dividing 60 seconds by the number of revolutions per minute, in this case 521).

So, t = 60/521 = 0.115 seconds

Therefore, α = 54.6/0.115 = 474.8 rad/s^2

Now, we can use this value of α to calculate the torque T using the formula:

T = I*α

Where I is the moment of inertia of the wheel. Assuming the wheel is a solid disk, the moment of inertia can be calculated using the formula:

I = 1/2 * m * r^2

Where m is the mass of the wheel (which is not given in the question, but can be assumed to be the same as the mass required to balance the wheel, 5g = 0.005kg) and r is the radius of the wheel (which is given as 0.33m).

So, I = 1/2 * 0.005 * (0.33)^2 = 0.00027225 kg m^2

Therefore, T = 0.00027225 * 474.8 = 0.1298 Nm

Now, we can plug this value of T into the formula for centrifugal force:

F = M * ω^2 * r

Where M is the mass of the attached mass (5g = 0.005kg), ω is the angular velocity (54.6 rad/s), and r is the radius of the wheel (0.33m).

So, F = 0.005 * (54.6)^2 * 0.33 = 9.521 N

Therefore, the centrifugal force acting on the rim of the wheel is approximately 9.521 N.