# [Mechanics] Calculate the braking torque

• hugo_faurand
In summary, the problem is that the rotational speed is an initial speed, not a constant speed, which complicates the calculations of the braking torque.
hugo_faurand
Homework Statement
We have a wheel spinning with a speed of 20rad/s.
The brake exerts a tangential force of 2073.6N and a normal force of 5760N.
(No radius given for the wheel)

1/Calculate the braking torque exerted by the braking force.

2/ Calculate how much rotations the wheel do before it stops

(3/ Which scaleswould be interesting to calculate ?)
Relevant Equations
Inertia of the wheel : 100m²/kg
μ=0.36
Hello ev
eryone !

I've got a problem in engineering class with a braking system (picture linked).
In the first part I calculated the friction force $$\vec{B}$$ with components :
T (following y axis) : 2073.6N
N (following x axis) : 5760N.

For the first question I struggle a lot because I haven't seen dynamics equations yet.
But on the web I saw that we can calculate the braking torque C with the equation :
$$C= J \times \alpha$$ alpha is the rotational acceleration. But our wheel is spinning at a constant speed so $$\alpha=0$$ and thus C=0 and that's strange. I found another formula : $$C= n \times N \times \mu \times r$$ r stands for the radius of the wheel, n is the number of pair of frictionning surfaces here 1 and N the normal force acting on the wheel.

This one is interesting because although we don't have the radius (I think that's an error from my teacher) we'll have a non-zero value.

For the question 2/ I use the kinetic energy theorem that tells us that the work of the system is equal to the difference of kinetic energies. So we have $$W = \frac{1}{2} J \omega_2^2 - \frac{1}{2} J \omega_1^2 = \frac{1}{2} J (\omega_2^2-\omega_1^2)$$
Whereas $$W=J\times \theta$$ thus we have $$\theta = \frac{W}{C}$$

Here is one solution for me and then we just have to divide the result by 2 pi to have the number of laps.

For the last question I've got no idea maybe the power...

So if you can help me because I'm not really sure of my work.
For the radius of the wheel maybe I can just take an arbitrary value, if the literal steps are good I think that's the most important.

One comment:
I think that the rotational speed is an initial speed, not a constant speed, as you state (##\alpha =0##). I say that because part of the problem is to determine when the wheel stops.

FactChecker said:
One comment:
I think that the rotational speed is an initial speed, not a constant speed, as you state (##\alpha =0##). I say that because part of the problem is to determine when the wheel stops.
That's right but we have no statement about the duration of the braking. So we can't determine the deceleration.

hugo_faurand said:
we can calculate the braking torque ##C## with the equation :$$C=J\times\alpha$$
Conversely, if we know ##C##, we can calculate the deceleration rate !

(assume you are right about '(I think that's an error from my teacher) ' and use the symbol ##r## )

Last edited:
I made all the calculations with a radius of 53mm and I've a torque of almost 39 N.m and it needs 4laps to stop. Is it coherent ?

(Otherwise spend good end of the year festivities !)

## 1. What is braking torque?

Braking torque is the force that is applied to a rotating object in order to slow it down or bring it to a complete stop.

## 2. How is braking torque calculated?

Braking torque is calculated by multiplying the coefficient of friction between the braking surface and the object being braked by the normal force applied to the object.

## 3. What factors affect braking torque?

The factors that affect braking torque include the coefficient of friction, the normal force, and the rotational speed of the object being braked.

## 4. How does the mass of an object affect braking torque?

The mass of an object does not directly affect braking torque, but it can affect the normal force applied to the object, which in turn affects the braking torque.

## 5. How can braking torque be increased?

Braking torque can be increased by increasing the coefficient of friction between the braking surface and the object, or by increasing the normal force applied to the object.

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