A 30 g ball rolls around a 40-cm-diameter L-shaped track, shown in FIGURE P8.53, at 60 rpm. What is the magnitude of the net force that the track exerts on the ball? Rolling friction can be neglected. Hint: The track exerts more than one force on the ball.

Question

Diagram showing forces acting on a ball in circular motion on a track.

Accepted Answer

Hi, everyone in this practice problem. We're being asked to determine the magnitude of the net force exerted by the track on the car. We will have an L shaped track using cardboard where the task object is going to be a small aluminum foil ball. The track has a diameter of 15 centimeter and the mass of the ball is going to be 30 g if the ball is moving or on the track at 25 R PM, ignoring rolling friction. We're being asked to determine the magnitude of the net force exerted by the track on the car. So the options given are 0.4, Newton, 0.6, Newton, 0.3 Newton and 0.8 Newton. And let's start off by listing and converting a bunch of the known values or the given values from the problem statement. So first we have the mass of the ball. So M which is going to be the mass of the ball is going to be 30 g or essentially it, it is going to be 30 multiplied by 10 to the power of negative three kg or essentially 0.3 kg. Next we have the diameter which is going to be 15 centimeter. And in this case, the radius will then be half of that which is 7.5 centimeter or essentially in si 0. m. The velocity is given to be 25 R PM. So I'm gonna list out 25 R PM, but I wanna change that to radiant per seconds. So 25 R PM will equals to 25 revolution per minute. Multiply that for every two pi radiance for every revolution, and we want to multiply that with one minute or every 60 seconds, that will actually give us the velocity of 2. radiant per seconds. Awesome. Looking at the figure given there are two compact points of the object with the track. So it is actually experiencing a normal force upward which is um represented by F N one here and a normal force inwards, which is represented by F N two here. Other than that, we obviously have the gravitational force pointing downwards countering the F N one. So what we wanna do to solve this problem is to actually by utilizing Newton's second law, I wanna utilize Newton second law and look at it both in the Z direction and also in the radial direction. So first, let's apply Newton's second law in the Z direction. I'm gonna represent that with Z there. So in the Z direction, sigma F C will actually equals to zero because it will not be experiencing any sort of movement in the Z direction. So in the Z direction or in the vertical direction, the two forces acting are F N one and F G. So Sigma F Z is going to be F N one minus F G with the convention of F N one pointing upwards or the upwards is going to be deposited in this case. So F N one minus F G will equals to zero. So F N one will actually equals to F G F G is just M multiplied by G. The mass is known to be 0.3 kg and the G is 9.81 m per second squared. So that will actually give us our F N one value which is going to be 0.3 Newton rounded up. Next, we want to actually find our F N two by applying the Newton second law in the radial direction or our direction. In this case, Sigma F R is going to equals two because there is some motion in the radial direction, it is not going to equals to zero and it will actually equals to M multiplied by AC ac being the cental acceleration. So in this case, uh the only force acting on to the radial direction is just F N two. So Sigma F R is going to just be F N two which will equals to M multiplied by AC. We wanna recall that AC is going to be represented by V squared divided by R. So in this case, we wanna substitute that into our equation. So that F N two will equals to M multiplied by V squared divided by R. The tangent shell velocity P can also then be represented by R multiplied by omega. So we want to substitute that back into our F N two equation so that our F N two will actually then equals to M multiplied by R multiplied by omega squared. Next, we can actually substitute all our, our known values because our velocity known here is actually representing angular philosophy. So this will actually equals to omega. And we also know what the mass and the radius is from the problem statement. So we can substitute the values in so that F N two will then equals to the mass which is 0.3 kg multiply that by the radius which is going to be 0.75 m multiplied that again by the uh angular velocity or the Omega which is going to be 2.62 radiance per second squared. That will give us an F N two value of 0.15 Newton. Awesome. So now that we have found R F N one and also R F N two, what we have left to do is to just find the magnitude of the net force asserted by the track on the car by taking the magnitude of F N one and F N two. So F net will actually equals to the square root of F N one squared plus F N two squared. We want to substitute our F N one and F N two. So that F net will equals to the square root of 0. Newton squared plus 0. Newton squared. They will leave us with the magnitude of F net being 0.3 Newton. So that will be the answer to this practice problem with an F net of 0.3 Newton or the magnitude of the net force exerted by the track on the car being 0.3 Newton, which will actually correspond with option C in our answer choices. So option C will be the answer to this practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topics and that will be it for this one. Thank you.

A 30 g ball rolls around a 40-cm-diameter L-shaped track, shown in FIGURE P8.53, at 60 rpm. What is the magnitude of the net force that the track exerts on the ball? Rolling friction can be neglected. Hint: The track exerts more than one force on the ball.

Verified Solution

Key Concepts

Centripetal Force

Net Force

Normal Force