An 8.0-cm-diameter, 400 g solid sphere is released from rest at the top of a 2.1-m-long, 25 incline. It rolls, without slipping, to the bottom..(a)What is the sphere’s angular velocity at the bottom of the incline?

Video transcript

Welcome back. Everyone in this problem, a solid disk with a diameter of 12 centimeters and a mass of six multiplied by 10 squared grams, starts from rest at the top of an inclined plane that has a length of 4 m along the incline and an angle of 35 degrees with the horizontal, calculate its angular speed when it rolls down to the base and assume that it rolls without slipping. It says it's 14 radians per second. B 35 radiance per second, C 66 radiant per second and D 91 radiance per second. Now, if we're going to find this angular speed, let's make sure we understand what's going on with the help of a diagram. So we have an inclined plane, ok. And on our inclined plane, we have a solid disk that we could draw at the top here with a diameter of 12 centimeters for our inclined plane that's 35 degrees to the horizontal. We also know that our inclined plane is 4 m, ok? So then we know that its vertical height is going to be a component of that 4 m. So let's call that edge and we know that it starts from rest at our plane and then rolls down to the base. So we can imagine this dotted line representing its direction and then this dotted disk representing our solid disk when it gets to the base of our incline. Ok. And at this point at the base of our incline, we want to figure out what its angular speed omega will be. Now, how can we find that? Well, first, let's make a note of the other information we have. We know that our disc, like we said, it has a diameter, let me write it under my diagram here. Diameter d of 12 centimeters or 0.12 m, which means that it will have a radius of 0.06 m, a half of 0.12. We also know it as a mass of basically 600 g, which is the same as 0.6 kg and the length of our incline, let's call that L is 4 m for an angle theater of 35 degrees. Now, for our solid disc, that rules in this scenario, we know that the principle of conservation of energy applies. In other words, our initial energy will be equal to our final energy. If we take our initial energy as the point where we were at the top of the inclined plane and our final energy when it gets to the base of the plane, then initially, we realized that we have a potential energy. So let, let's make a note here by the conservation of energy. That means that our initial energy which is solely potential energy because it's our problem says it starts from rest. OK. So our initial energy which is just potential is going to be equal to our final energy. Now, when we get to the base of the hill, our height will be zero. There's no potential energy, we only have kinetic energy but not that kinetic energy is going to be made up of translational energy. So let's call that ket and rotational energy. So let's call that K er so our initial potential energy will be equal to the sum of both those kinetic energies. Now, what do we know about each of our energies? Well, recall, OK, let me just fix this line here. Recall that potential gravitational energy is equal to the mass multiplied by gravity multiplied by a perpendicular height. OK? Or a vertical height. Rather, our translational kinetic energy is equal to half of MV squared. And our rotational kinetic energy is equal to half of I multiplied by omega squared. OK. So with that in mind, then that tells us that MGH is going to be equal to M sorry half of MV squared. OK, plus half of I omega squared. Now remember we're solving for omega or angular speed. So we will need to first figure out what I is or moment of inertia. And we'll also need to rewrite our translational energy in terms of our angular speed to eventually help us solve for our angular speed. So how can we do that? What information do we have? Well, we also know that when it comes to a solid disk, its moment of inertia is given by half of Mr squared. OK. And we also know that our speed or linear speed is equal to our angular speed multiplied by the radius of our object. In this case are. So if we substitute both of those, then that tells us that MGH is going to be equal to a half of M multiplied by Omega R squared plus a half of our moment of inertia, which is a half Mr squared, OK? Multiplied by omega squared. So, well, let me write it beside it because that makes it seem like it's above it. So that's a half of Mr squared. OK? Multiplied by Omega squared. We're almost there. No, we just need to figure out what our value for H is. Let's go back to our diagram like we said, is going to be the vertical component of our length of 4 m. OK. So since it's opposite to a 35 degree angle, we can tell them that H is going to be equal to L multiplied by the sine of theta. In this case, that's gonna be 4 m multiplied by the sine of 35 degrees. No, I'm not gonna solve yet. I'm just going to put it into our problem. OK. So again, when we do that, and when we expand our bracket here, that's going to be MG multiplied by L sine theta being equal to half of M multiplied by omega squared R squared plus a quarter of Mr squared omega squared. Now, we can start to simplify first, we can cancel M from all of our terms. OK. Then we can uh factor out omega squared here because remember that's what we're trying to solve for. So when we do that, we get GL Sine theta being equal to omega squared multiplied by half R squared plus a quarter of R squared, which conveniently enough that can simplify to three quarters of R squared. And that's, that's, that's pretty convenient. OK. So that's three quarters of R squared. No, that we have that we can go ahead and solve for Omega squared because no, that means Omega squared. If we divide, if we divide both sides by three quarters of our squared, that means it's going to be equal to, it's gonna be equal to four, multiplied by GL sine theta. OK? Divided by three multiplied by R squared. No, we're almost there. No, we just need to go ahead and square root both sides. OK. So when we do that, we get Omega then the angular speed to be equal to the square root of four GL sine theta divided by three R squared, let's go ahead and substitute our values. So that's going to be four multiplied by the acceleration due to gravity. We can take that to be uh 9.8 m per second squared, multiplied by our length of 4 m multiplied by the s of 35 degrees. And that's all divided by three. OK. Three multiplied by a radius of 0.06 m squared. So to find our answer, we just need to find a square root of this expression. Now, when you go ahead and do that, OK, then you should find that you get Omega to be equal to 91.25 radiance per second. Remember all of our answers are into two significant figures. So if we do the same, then we get our angular speed to be approximately equal to 91 radiance per second. If we go back to our answer choices that tells us, then that d is the correct answer. Thanks a lot for watching everyone. I hope this video helped.