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Question 2 To work out how hard the pads must push against the disk, let's start off by working out the friction force caused by friction between the tyres and the road that allows me to slow down in 10m distance. Full speed for me on my bike is about 40 km/hr which is roughly 10 m/s. How much backwards friction force is needed to slow me down over 10m distance? We could work this out in two ways. One would be to calculate the acceleration needed to stop in this distance, and then use " = ma to work out the force. Or use , = u' + 20s if you happen to have memorised that equation (I haven't). I would probably use work to do this. Work is force times distance-you know the distance and want to work out the force. And the (negative) work done must equal the kinetic energy removed. If my bike and I have a combined mass of around 100 kg, estimate the stopping force required to stop me in 10m (in N). Answer: Check Let's assume that half this backwards force came from each wheel. Disk Brakes This force applies an anti-clockwise torque to each wheel. If we assume the wheels have a radius yz ~ 0).7m. what is the torque applied by this friction force to each wheel? (Remember that only half the total braking force is applied to each individual wheel). Answer: CheckThis torque is trying to make the wheels spin faster! But in fact, they spin slower as you slow to a halt, so there must be another bigger torque, from the brakes, that is acting in a clockwise direction. We know that this brake-pad torque must be bigger than the friction torque. But how much bigger? Much bigger, or just slightly bigger? This depends on which is the bigger task: Overcoming the angular momentum of the wheel. Slowing down the bike and rider. Which is bigger? You could calculate the angular momentum of the wheel, and work out the extra torque needed to slow it to a stop, over 10m. But imagine spinning the wheel when it is raised off the ground. How hard is it to stop then, compared to when you are riding it? Based on this, which of the following is true? Select one: Select one: O The brake torque is MUCH more than the friction torque. O The brake torque will be only slightly more than the friction torque. Check It's pretty easy to use the brakes to stop a wheel spinning when it's not touching the ground, so that tells us that the torque applied by the brakes is only slightly more than the friction torque. So let's set them equal. If the brake-pads are " = 7 cm from the axle, what clockwise force must the brake pads apply to the wheel? Answer: Check The force must come from friction between the brake pads and the disk. The pads push on both sides, and there will be a friction force on both sides (each supplying half the force). Coefficient of friction is probably pretty high (around 1). So how much force does each side of the brake pad apply to the disk? Answer: Check