Andrew B. answered • 05/02/24
PhD in Applied Physics with 10+ Years of College Teaching Experience
Here is the text of the problem description:
"A solid sphere of radius 36 cm is positioned at the top of an incline that makes 27° with the horizontal. This initial position of the sphere is a vertical distance 3.1 m above its position when at the bottom of the incline. The sphere is released and moves down the incline. Find the speed of the sphere when it reaches the bottom of the incline if it rolls without slipping. The acceleration of gravity is 9.8 m/s2 and the moment of inertia of a sphere with respect to an axis through its center is 2/5 M R2." There is also a very helpful image of the incline with the values given.
The first thing to note is that it is Rolling without Slipping. This means the following things are true:
1) The linear and rotational motion are linked by the radius, so linear velocity = the radius times the angular velocity, and so forth.
2) All the energy is mechanical. In other words, no energy is transferred to heat as it slides down, so the total energy = linear kinetic energy + rotational kinetic energy + gravitational potential energy. Furthermore, the kinetic energy is split between linear and rotational based on the Moment of Inertia.
The second thing to note is that we are dealing with the motion down an incline plane where we start from rest at the top and want the speed at the bottom. This lends itself to using the Conservation of Energy to solve the problem. This means that the total energy at the top must equal the total energy at the bottom.
At the top: The sphere is at rest, so it has no kinetic energy (linear or rotational). Therefore, all of the energy is gravitational potential energy: V = mgh. Note that we were not given the mass in this problem, so let's hold off on putting in numbers until the very end (this is good practice anyways). With that, we have:
Total Energy = Linear K + Rotational K + Potential = 0 + 0 + mgh
Total Energy = mgh
At the bottom: Since it's at the bottom now, the height is zero, so the potential energy is zero. This means that all of the energy is kinetic (linear and rotational):
Total E = Linear K + Rotational K + Potential = 1/2 m v2 + 1/2 I ω2 + 0.
Using the Conservation of Energy:
The Total Energy at Top = The Total Energy at the Bottom
mgh = 1/2 m v2 + 1/2 I ω2
Rolling without Slipping: Connecting the Linear and Rotational Motion:
Next, let's insert our equation for the moment of inertia, given as 2/5 M R^2.
Also, linear and rotational motion are linked via the radius for rolling without slipping, so the linear speed = radius times the angular speed: v = r ω and ω = v / r.
Inserting those statements into the above energy equation, we have
mgh = 1/2 m v2 + 1/2 • (2/5 m r^2) • (v/r)2
mgh = 1/2 m v2 + 1/5 m v2
Note how the radius cancels out in the second term, and the mass is common in each term. Combining the 1/2 and 1/5, we have:
gh = 0.7 v2
v = sqrt(gh / 0.7) = sqrt(9.8 m/s2 • 3.1 m / 0.7)
v = 6.59 m/s
As a side note, this problem illustrates that for rolling without slipping, how the kinetic energy is split between linear and rotational is purely based on the shape of the object (the A in I = A m r2, which was 2/5 in this case).
