. How is it that the torques cancel when you turn a bicycle?
During a turn, you lean the bicycle into the turn. For example, when you turn left, you lean the top of the bicycle toward the left. The result is that you (and the bicycle) experience two torques. First, the support force from the ground tries to rotate you one direction—it tries to make your head go left and your feet go right. Second, friction from the ground, which is making you and the bicycle accelerate toward the left as part of the turn, tries to rotate you in the opposite direction—it tries to make your head go right and your feet go left. These two torques will cancel one another if you are leaning just the right amount. As a result, the bicycle doesn't undergo angular acceleration and you don't tip over.
. Is there a difference in the types of handles of a bike? On some bikes, there is the (upright) handlebar and on some the (drop) handlebar. Is there a purpose?
The shape of the handlebar determines your riding position. The upright position is generally more comfortable but, by sitting you upright, it increases the pressure drag you experience. Drop handlebars lower your body and make you more aerodynamic, but that position isn't as comfortable.
. Please explain the advantage of air tires with less mass.
The rim of the wheel travels at a different speed from the rest of the bicycle. The top of the wheel heads forward faster than the bicycle, while the bottom of the wheel heads forward more slowly than the bicycle. But because kinetic energy is proportional to the square of speed, the increase in the top of the wheel's energy caused by its increased speed more than makes up for the decrease in the bottom of the wheel's energy caused by its reduced speed. The overall result is that the wheel rim has twice as much kinetic energy as it would have if it were simply sliding forward without turning. This fact is important because it means that you want as little mass in the wheel rim as possible. Every kilogram there counts double when you are trying to start up from rest. By putting air inside the tire, rather than rubber, you reduce the mass at the wheel rim and make the bicycle easier to start.
. Why do the front brakes of a bike provide more braking power than the rear brakes, assuming both are applied with equal pressure?
When you apply brakes on a bicycle, you make it harder for the wheels to turn. The ground must then exert backward frictional forces on the wheels to keep them turning and it is these backward frictional forces that slow the bicycle's forward motion. But the forces that the ground exerts on the bottoms of the wheels also produces a torque on the bicycle about its center of mass—the whole bicycle has a tendency to begin rotating. Fortunately, the bicycle rarely actually rotates—if it did, you would fly forward over the front wheel of the bicycle. But this tendency to rotate during braking pushes the bicycle's front wheel downward onto the pavement and lifts the bicycle's back wheel upward off the pavement. The added pressure between the front wheel and the pavement improves traction there and makes the front wheel particularly effective for braking. The loss of pressure between the back wheel and the pavement reduces traction there and makes the back wheel particularly ineffective for braking. In fact, it's easy to begin fishtailing as the rear wheel loses traction completely.
. Why is it easier for you to make sharp turns more quickly when your center of gravity is over the handle bar?
The force that causes you to turn is friction between the front wheel and the ground. When you turn left, friction pushes the front wheel left and you turn left. By putting all of your weight over the front wheel, you accomplish two things. First, you increase the maximum static frictional force between the ground and the front wheel. You push them together harder so that they are less likely to slide (skid). Second, you make it easier for that sideways friction force to accelerate you; the force acts closer to your body and more directly on you. There are fewer torques on the bicycle that might cause it to skid about on either the front or rear wheel.
. If you apply the brakes while making, say, a left turn on a motorcycle, the motorcycle will tend to "stand up." That is, it will tend to fight the lean you make into the turn. Why?
When you turn left, you are accelerating toward the left and your velocity is changing toward the left. This leftward acceleration requires a leftward force and that force is supplied by friction between the ground and the motorcycle's wheels—the ground pushes the wheels toward the left. However, this leftward force on the wheels also exerts a torque (a twist) on the motorcycle about it's own natural point of rotation—its center of mass. As the ground pushes the wheels toward the left, the motorcycle tends to begin rotating. In this rotation, the wheels begin moving toward the left and the driver's head begins moving toward the right—the motorcycle "stands up"! Actually, if you lean far enough to the left as you turn, an opposing torque due to the upward force that the road exerts on the wheels will balance the first torque and your motorcycle will experience no net torque—it won't stand up at all. On a high-speed turn, you must lean quite a bit to avoid the "standing up" problem, which is why motorcycle racers practically touch the ground as they turn.
. Why is it harder to balance on a stationary motorbike compared to a moving one? Is it a gyroscopic effect on the wheels? — DF, Morley Perth, Australia
As you suspect, gyroscopic effects do play a role. Because it has only two wheels, a motorbike is inherently unstable. When it's stationary, it is only in equilibrium—that is it experiences no net force or torque—when it's perfectly upright. The slightest tip causes it to fall over. You must be very careful and agile to keep it balanced. A physicist would say that the motorbike is statically unstable or that it has an unstable static equilibrium.
For the motorbike to remain upright, you must keep the overall center of gravity (yours and the motorbike's) directly above the wheels (actually the line formed by their contact points on the ground). That's very hard to do when the motorbike is stationary. But when the motorbike is heading forward, it naturally steers itself under the center of gravity. If the motorbike begins to tip to one side, its front wheel automatically steers in the direction of the tip and the forward moving motorbike soon drives its wheels back under the center of gravity. This automatic steering is due to both gyroscopic precession in its spinning front wheel and to the shape and angle of the front wheel fork. If you hold the motorbike (or a bicycle) off the ground, spin its front wheel the right direction, and then tip the motorbike, you'll see its wheel turn toward the direction of the tip because of gyroscopic precession. If you return the motorbike to the ground and then tip it to one side, you'll see that its wheel will automatically turn toward that side because of the fork shape.
With both effects helping the motorbike steer under the center of gravity, the moving motorbike is very stable. A physicist would say that it is dynamically stable. Everything I've said also applies to bicycles and was pointed out by British physicist David Jones in 1970. Bicycles are so dynamically stable that almost anyone can ride them without hands and not tip over!
. What are the relative efficiencies and reasons for power losses in sprocket and chain drives, rubber cogged belt drives, pulley drives, and gear drives? — RA, Montreal, Quebec
The only power loss mechanisms I can think of in each case are sliding friction and vibration. The drive system most likely to experience substantial sliding friction is a pulley (or smooth belt) drive. If the belt slips as the pulleys turn, the belt will do work against the force of sliding friction and that work will be converted into thermal energy. But, as one of my readers points out, if the belt is properly tightened, has an adequate coefficient of friction to prevent slipping, and has a high tensile strength so that it doesn't creep across the pulley surface, then it can operate with very little power loss.
In the other drive systems, there is no possibility of slippage so that any power loss that occurs must be due to internal sliding friction within the components, or from vibrations. Flexing a chain involves some internal sliding friction and wastes some power. I suppose this could be minimized with careful chain construction and I wouldn't be surprised if large change drive systems placed bearings in the chain links to eliminate sliding friction altogether. Flexing a rubber-cogged belt also involves some molecular friction within the belt material so it wastes some power. I'm not sure which system is more efficient, the chain drive or the cogged belt drive. Finally, the gear drive is the least likely to waste significant energy. The only sliding friction that occurs is between the gear teeth. If the teeth are designed well and cut carefully, they should slide very little. In that case, the only significant power loss would be through vibrations. If everything is carefully mounted to prevent vibrations, there should be very little power loss in a gear drive.
. Can you explain gyroscopic precession? — BW, Newport, RI
When a gyroscope is spinning rapidly, it has a large amount of a conserved physical quantity called angular momentum. Angular momentum is a special measure of rotational motion that can't be created or destroyed—it can only be transferred between objects. As long as nothing tries to transfer angular momentum to or from the spinning gyroscope, it will continue to spin at a steady pace about a fixed axis in space. But when an external torque (a twist) is exerted on the gyroscope, a transfer of angular momentum takes place. The gyroscope's rate of rotation or its axis of rotation begins to change so that its angular momentum changes. If you apply a twist to the gyroscope around its axis of rotation, it will either spin faster or slower, depending on which way you twist it. But if you twist the gyroscope about a different axis, its axis of rotation will shift—the gyroscope will undergo precession. The direction of this precession depends on how you apply the twist and tends to be very non-intuitive.
. How do bicycle shocks and suspension affect the performance of a bicycle? — D
When the wheels of a bicycle are attached directly to the frame of a bicycle, the wheels and frame must move together. When one of the wheels hits a bump, both that wheel and the frame must accelerate upward together. When this happens, the bump exerts a huge upward force on the wheel and everything, including the unfortunate rider, experiences a sudden upward acceleration. A sudden jolt of this sort is unpleasant—the seat of the bicycle pushes upward violently on the rider and the rider feels large forces throughout his or her body. Each body part pushes upward on the body part above it so that everything leaps upward.
To reduce the upward acceleration that the rider experiences, the direct connection between the bicycle wheels and the frame can be replaced by a spring suspension. When the wheel of a bicycle with a spring suspension encounters a bump, the springs compress and the force on the frame and rider is much smaller. The rider still accelerates upward, but not as rapidly as the wheel and without the abrupt jolt of a suspensionless bicycle. In fact, by the time the rider has begun to rise much, the wheel will probably have rolled back off the bump and the spring will return to its original shape. Overall, the rider will barely move at all and will hardly notice the bump.
But a spring suspension isn't perfect by itself. Suppose that the bicycle rolled over a curb and onto a sidewalk. This bump doesn't end—the pavement level rises permanently. When the wheel hits the curb, it rises suddenly and compresses the spring. But since the wheel never drops back to its original height, the only way for the spring to decompress back to its original shape is for the frame and rider to rise. And that's what happens. But the frame and rider don't stop moving once the spring has reached its original shape. They have upward momentum and they continuing rising. The spring begins to stretch upward now. Eventually the frame and rider stop rising and begin to descend again, but they continue to bounce up and down as though they were on a pogo stick. In effect, they are on a pogo stick. When a spring is compressed or stretch, it stores energy. If there is nothing to get rid of the energy stored in the bicycle's compressed or stretched spring, the frame and rider will continue to bounce up and down indefinitely.
To stop the bouncing (and prevent most of it in the first place), a bicycle with a spring suspension also has shock absorbers. These devices waste energy whenever the wheel and frame move relative to one another. Whether the spring is compressing or stretching, the shock absorber extracts energy from the wheel, frame, and spring, and turns that energy into thermal energy. As a result, the frame and rider don't bounce significantly after the wheel rides up and onto the curb. Similar issues occur in cars, where shock absorbers damp out the bouncing that can occur because the car body is suspended above the wheels on springs.