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Tutor profile: Kevin R.

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Kevin R.
Graduate Research Assistant at Missouri S&T
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Questions

Subject: Physics (Newtonian Mechanics)

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Question:

Consider an inclined ramp where at the top you have three objects that you intend to roll down the ramp simultaneously. Each object will roll without slipping. One is a solid sphere, one a solid cylinder, and the last one a simple hoop. You plan the "race" them and see which one "wins". You are also interested in the order in which they arrive. Find the order in which the object reach the bottom if you release them from rest at the same time.

Inactive
Kevin R.
Answer:

Recall that for simple rolling objects the moment of inertia is a measure of how easy it is to roll something. It is analogous to mass and how difficult it is to move. The moment of inertia is determined by how the mass of the object is distributed; objects with mass farther away from the center have a higher moment of inertia than those with mass close to the center. With this in mind we can predict the order of the objects. Qualitatively, the hoop will be last and the sphere will be first. We can see this by noticing that a hoop as all its mass on its "edge" while a sphere has most of its mass close to its center. The cylinder will fall in between. Consulting a table for moments of inertia you will find that for the hoop it is m*R^2, for the solid cylinder it is (1/2)*m*R^2, and the solid sphere is (2/5)*m*R^2. Here m is the mass of the object and R is the radius of the object. The sizes of the moments of inertia for each object falls in line with our prediction for a given m and given R. For a quantitative answer we can use energy considerations to solve it. Let the height of the inclined ramp be h. Then for an object that we release from rest, the speed of the object as it rolls down the incline can be found with: m*g*h = (1/2)*m*v^2 + (1/2)*I*w^2. This equation is an energy balance where the left hand side has the gravitational potential energy of the object, and the right hand side has the linear and rotational kinetic energies respectively. m is still mass, g is the acceleration due to gravity (~9.8m/s^2), v is the speed of the object after moving down a distance h, I is the moment of inertia, and w is the angular speed of the rolling object. Given that the objects are rolling without slipping we can use: w = v / R. For most simple objects, like those in this problem, I = K*m*R^2 where K is some constant (i.e 1/2, 1, or 2/5), m and R are the same as above. Plugging these two relations into our energy equation and solving for v we find: v = Sqrt( 2*g*h / (1 + K) ). From this we see that objects with smaller K will have the faster final speed. This will be true for all distances h as the objects roll down the incline. Consulting the constants K from above, we see that our qualitative prediction was correct! Notice the important secondary result is that the mass of the object and it's radius (a measure of its size) does not show up at all in our final result. The result of the race will be the same no matter the masses and sizes of the objects rolled. Only the geometry of the objects will determine the winner!

Subject: Physics (Electricity and Magnetism)

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Question:

Consider a system of two point charges separated by a fixed distance. Initially, you measure the magnitude of the force between them to be Fo. Now you increase one charge by 10% and the other you reduce by 10% while keeping the distance between them fixed. Will the new magnitude of the force, F1, between them remain the same or change? If it changes, then find the ratio between the new force and the old.

Inactive
Kevin R.
Answer:

Recall that the equation of the force between two point charges is given by Coulomb's Law: F = k*q1*q2 / r^2 where k is a fixed constant, q1 and q2 are the charges, and r is the distance between them. The changes in the charges correspond to q1 -> 0.9*q1 and q2 -> 1.1*q2. After substituting these relations in we find: F1 = k*0.9*1.1*q1*q2 / r^2 = 0.99*Fo. Solving for F1 / Fo we find the ratio to be 0.99. The force will always decrease no matter what q1 and q2 were initially!

Subject: Physics

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Question:

Consider a simple pendulum where a sphere is hung from the ceiling by a light string. The string has negligible mass and it's length is much larger than the size of the sphere so that you may treat the system as a Simple Pendulum. The sphere is is filled with sand and has a tiny hole at the bottom that will slowly leak the sand out. Given that the pendulum has an initial period of To, describe qualitatively how the period will change in time as the sand leaks out.

Inactive
Kevin R.
Answer:

Recall that in the study of simple pendulums the period of oscillation only depends on the distance from the pivot to the center of mass of the hanging weight. The relation is T = C*Sqrt(L), where C is a constant that isn't important to the problem since the answer is qualitative, T is the period, and L is the distance from the pivot to the center of mass. Note from this relation the mass of the hanging weight also does not matter and it is not contained in C. As the sand leaks out of the bottom of the sphere the center of mass will shift downwards. Initially the period of the pendulum will increase. As the sand continues to leak out it will eventually have less mass than the sphere itself, the center of mass will shift back upwards, and the period will then decrease. Once all the sand is out the period will return to To.

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