# Math Help - Angular deceleration of wheels.

1. ## Angular deceleration of wheels.

Right, so I have been stuck on this question for a while now but not sure where to start. I'm almost pulling my hair out, any guidance or pointers would be appreciated.

I need to calculate the angular deceleration of wheels of a lorry. It needs to stop in 0.5km using constant braking force.

It's weight is 5 tonnes and velocity is 80 km h-1. The wheels are 1.5m in diameter.

So far I've worked out the deceleration of the lorry (-0.494 ms-2), the braking force required (-2.4kN), the time taken to come to rest (2024.29), and the initial angular velocity of the wheels (29.63 rad/s).

Just can't get my head around the "angular deceleration" part. Can someone please tell me where to start.

Regards,

Becca.

2. You can use the same formula as with linear speed acceleration/deceleration.

$v^2 = u^2 + 2as$

v = 0,
u = initial angular velocity
a = acceleration or deceleration
s = angular displacement

3. Hi, thanks for the reply.

Could I use alpha=dw/dt?

4. Originally Posted by Unknown008
You can use the same formula as with linear speed acceleration/deceleration.

$v^2 = u^2 + 2as$

v = 0,
u = initial angular velocity
a = acceleration or deceleration
s = angular displacement
Originally Posted by becca

Could I use alpha=dw/dt?
You do not have d $\omega$/dt in order to do this. However you have the initial angular speed (you know the radius of the wheel and the linear speed, so $v_0 = r \omega _0$ ). You know the final angular speed, and you know the angular distance the wheel has gone through. ( $s = r \theta$ ) So you can use the angular analogue of Unknonw008's equation: $\omega ^2 = \omega _0^2 + 2 \alpha \theta$ .

By the way, the two equations $v_0 = r \omega _0$ and $s = r \theta$ are known as the "rolling without slipping" conditions. There is one more: $a = r \alpha$. Whenever a wheel is moving without slipping these relations relate the linear variables (s, v, and a) to the rotating variables ( $\theta$, $\omega$, and $\alpha$.)

-Dan

5. Hi Dan,

So using your formula for acceleration I get:

α=(ω_0^2- ω^2)/θ

α=(29.63-0)/6.28