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Consider the the front wheels of a roller coaster cart. Presumably, when the roller coaster in in a straight section the left and right wheels are moving at the same speed. But when the roller coaster enters a turn, the wheels on the outside of the turn must move faster than the wheels on the inside, since the outer wheels need to travel a longer distance (larger radius) than the inside wheels.

What force causes the outer wheels to speed up (or the inner wheels to slow down)?

See the picture below. At (A) the wheels are turning the same speed in the straight section of track. At (B) the cart is just about to enter a turn, still going straight. At (C) the cart is in a turn, and the outer (left) wheel must be spinning faster than the inner (right) wheel. What force was applied between (B) and (C) to make this happen?

roller_coaster_turn

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  • $\begingroup$ The body of your question and the title are inconsistent. What causes the cart to turn and what causes the wheels to turn at different speeds don't have to be related at all. Please edit your question to ask just one thing, with the title reflecting the body of the question, or visa-versa. $\endgroup$
    – TimWescott
    Commented Jun 27, 2023 at 22:51
  • $\begingroup$ Can you explain how they don't have to be related? The title is what I am truly interested in. An initial draft of this asked about the outer side of the axle moving faster than the inner side, and in a cart with a rigid construction this seems sufficient to understand how the whole cart turns. But asking about the axle seemed less intuitive than the wheels. But now I'm curious why those aren't essentially the same question and what different forces would cause the axle/cart to move thats different from the wheels. $\endgroup$
    – mv3
    Commented Jun 28, 2023 at 12:29

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At (C) the cart is in a turn, and the outer (left) wheel must be spinning faster than the inner (right) wheel.

... or the wheels are slipping (skidding) as they move along the track.

What force causes the outer wheels to speed up (or the inner wheels to slow down)?

The interaction with the track, which depends heavily on the specifics of the construction.

For a real railroad, the wheels do not have a constant diameter. The turn allows the outside wheel to ride on a larger diameter, so it can cover more distance as it turns.

For an amusement ride, efficiency may not matter much, but there must be some mechanism to bring lateral force on the cart as it reaches the side of the track. As the cart reaches the turn, the lateral force from the track will push right on the front of the cart. This creates a torque that tends to turn the entire cart to the right.

For a simple cart (solid axle, simple wheels), then the wheels may simply skid during a turn. The outer wheel covers more ground than the inner wheel, but turns the same amount. This will cause increased drag, but may not matter much for a short ride.

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  • $\begingroup$ I thought about the wheels slipping, but even in this case, the axle still needs to twist and so the outer part of the axle should somehow be moving faster (is this correct?) I think the part about the lateral force providing a torque is what I'm mentally missing or not super clear about. Thinking about a hotwheels car on a track, it goes around a turn by just bumping the wall - I guess that bump creates a torque similar to how the later forces of being constrained to a track on a roller coaster would. Is that correct? $\endgroup$
    – mv3
    Commented Jun 27, 2023 at 21:17
  • $\begingroup$ Yep. For hotwheels, they don't even need a track. The car itself can bump into something and that provides the lateral force. $\endgroup$
    – BowlOfRed
    Commented Jun 27, 2023 at 21:46
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The only forces acting in this scenario are gravity, friction, and the normal force from the track. Gravity and the normal force both act perpendicular to the direction of motion and through the axis of rotation, leaving only friction as the only possible culprit for changing the angular speed of the wheels. Intuitively, this makes sense - without friction, the spin rate of the wheels would be entirely irrelevant, as the car would simply slide along the track no matter the speed of the wheels. Without friction, the wheels don't change speed when rounding the curve, but they do when friction is present - friction is responsible for changing the spin rate of the wheels.

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What force causes a roller coaster to turn in a track?

The next time you're close to a roller coaster, look at the shape of the track's rails and the shape and arrangement of the wheels.

Or look at this diagram, found here:

enter image description here

There are some important things to notice here. The relevant one for this part of your question is that the roller coaster wheel and axle assembly is constrained to ride on the track in every direction except for forward and backward motion. The wheels can't push down through the track, so the track supports the cart when it is still. The wheels can't pull up through the track, so the track holds the cart down when it is trying to fly off. The wheels cannot push into the track from either side, so the cart is constrained to stay centered on the track.

When the roller coaster cart is moving and the track turns a corner, the cart has to follow the track. The side force to make this happen comes from the side wheels pressing on the side of the track, on the outside of the corner.

What force causes the outer wheels to speed up (or the inner wheels to slow down)?

Because the wheels each have their own independent bearings and are not connected by a solid axle, they are free to rotate at a speed dictated by their contact with the track. The force that causes the outer wheels to speed up and the inner wheels to slow down is the frictional force of the wheels riding on the track.

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