Help understanding circular motion

[Isiudor]

Homework Statement

A skateboarder with mass 61kg sliding down a circular ramp reaches a speed of 7.8 m/s at the bottom of the ramp
what is the net force acting on the skateboarder at that point?

Homework Equations

N = mg
Fr = m * v^2/r

The Attempt at a Solution

In my opinion the only force acting on the skateboarder at the bottom of the ramp is the NORMAL force which is m*g = 597N

but since the answer in the book is 1127 it seems it takes into account some fictional centrepintal force which is m * v^2/r = 530

530+597 = 1127

however as far as I can tell there is no actual force pulling him inward towards the center of the ramp but and the only reason he stays in a "circular motion" at that point is the solid ramp preventing his escape, now if we were at the top of the ramp it would have been a different matter.

I also have a second question deriving from this one, if I were asked what are the net forces acting on this skateboarder not at the bottom or the top, but rather at an angle, say 30 degrees from center where 0 is the bottom.

infact if someone could explain all kinds of degrees,say... 30, 90 and 130? how do I even approach that?

 

[Doc Al]

In my opinion the only force acting on the skateboarder at the bottom of the ramp is the NORMAL force which is m*g = 597N

m*g is the weight of the skateboarder, not the normal force.

but since the answer in the book is 1127 it seems it takes into account some fictional centrepintal force which is m * v^2/r = 530

Are you sure the problem asks for the net force? The net force is given by Newton's 2nd law, which in this case gives: F_net = m*v^2/r.

The 1127 N force is the upward normal force exerted by the ramp on the skateboarder. Nothing fictional about it! (I assume the radius of the ramp is 7 m?)

530+597 = 1127

however as far as I can tell there is no actual force pulling him inward towards the center of the ramp but and the only reason he stays in a "circular motion" at that point is the solid ramp preventing his escape, now if we were at the top of the ramp it would have been a different matter.

The way that the ramp forces the skater into circular motion is by exerting a very real force!

I also have a second question deriving from this one, if I were asked what are the net forces acting on this skateboarder not at the bottom or the top, but rather at an angle, say 30 degrees from center where 0 is the bottom.

You'll find the net force using Newton's 2nd law in all cases.

 

[Isiudor]

well doc, thanks for the fast reply, good to see you are still around.

anyway, yes the radius is 7m and the question does ask for the total force acting upwards.

so you're saying that ontop of the mg force acted back on the skateboarder due to it's weight(597), the ramp exerts additional force to keep it in circular motion(530) hence the total force acted upwards is 1127? is this true cause you wrote that fnet is only the circular force which is 530, yet the anwer in my pdf says the total upward force is 1127N.

 

[Doc Al]

so you're saying that ontop of the mg force acted back on the skateboarder due to it's weight(597), the ramp exerts additional force to keep it in circular motion(530) hence the total force acted upwards is 1127?

No. The weight pulls down with a force of 597 N while the ramp pushes up with a force of 1127 N. The net force, given by mv^2/r, is 1127 - 597 = 530 N upward. This net force is called the 'centripetal' force.

is this true cause you wrote that fnet is only the circular force which is 530, yet the anwer in my pdf says the total upward force is 1127N.

They must mean the total force exerted by the ramp, not the net force acting on the skateboarder.

 

[rwisz]

I can offer some clarification and explanations for your questions about circular motion.

First, let's define circular motion. It is a type of motion where an object moves along a circular path, constantly changing its direction but maintaining a constant speed. In this case, the skateboarder is moving along a circular ramp, so we can consider it as an example of circular motion.

Now, let's break down the forces acting on the skateboarder at the bottom of the ramp. As you correctly pointed out, the normal force (N) is the force exerted by the ramp on the skateboarder, perpendicular to the surface. This is the only force acting on the skateboarder in the vertical direction. However, there is also a horizontal force acting on the skateboarder, which is the centripetal force (Fc). This force is directed towards the center of the circular path and is responsible for keeping the skateboarder in circular motion. It is given by the equation Fc = m * v^2/r, where m is the mass of the skateboarder, v is the speed, and r is the radius of the circular path.

So, to calculate the net force (Fnet) acting on the skateboarder at the bottom of the ramp, we can use the equation Fnet = N + Fc. This is because both forces act in different directions, so we need to add them together to get the total force.

Now, let's address your second question about the net forces at different angles on the ramp. To approach this, we can use the same equation Fnet = N + Fc, but we need to consider the angles at which the forces act. At 30 degrees, the normal force will still act perpendicular to the surface, but the centripetal force will now act at an angle towards the center of the circular path. This means we need to use some trigonometry to calculate the horizontal component of the centripetal force, which can then be added to the normal force to get the net force.

At 90 degrees, the skateboarder is at the top of the ramp, so the normal force will act downwards and the centripetal force will act towards the center of the circular path. At 130 degrees, the normal force will act at an angle downwards and the centripetal force will act at an angle towards the center of the circular path. Again, we can use trigonometry to calculate the horizontal component of the