# Angular deceleration of wheels.

• Feb 14th 2011, 08:31 AM
becca
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.
• Feb 14th 2011, 10:27 AM
Unknown008
You can use the same formula as with linear speed acceleration/deceleration.

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

v = 0,
u = initial angular velocity
a = acceleration or deceleration
s = angular displacement
• Feb 14th 2011, 10:59 AM
becca

Could I use alpha=dw/dt?
• Feb 14th 2011, 11:37 AM
topsquark
Quote:

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

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

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

Quote:

Originally Posted by becca

Could I use alpha=dw/dt?

You do not have d $\displaystyle \omega$/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 $\displaystyle v_0 = r \omega _0$ ). You know the final angular speed, and you know the angular distance the wheel has gone through. ( $\displaystyle s = r \theta$ ) So you can use the angular analogue of Unknonw008's equation: $\displaystyle \omega ^2 = \omega _0^2 + 2 \alpha \theta$ .

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

-Dan
• Feb 15th 2011, 09:05 AM
becca
Hi Dan,

So using your formula for acceleration I get:

α=(ω_0^2- ω^2)/θ

α=(29.63-0)/6.28