Drawing Free Body Diagrams with torque

[juggalomike]

Homework Statement

This is a review problem i was given for my final exam tommorow, I am stuck on 1 problem, even with the answer key I am not understanding it.(also sorry but the picture won't copy from MSword, but its a cylinder rolling up a hill)

(B) When the lander hits the surface, it eventually stops bouncing and finds itself rolling up the side of a crater wall inclined at an angle of theta, still wrapped in its airbags.

To a good first approximation, the lander is a uniform cylinder of mass M and radius R. The airbags can be thought of as forming a thin shelled sphere of radius r and mass m around the lander. Assume that the lander is rolling up the steepest incline possible without slipping. It terms of the given quantities, show a good step-by-step method to find out what the acceleration of the lander will be. Use g of Mars for the acceleration due to gravity on the surface of Mars.

Draw FBD
∑F ⃗ =ma ⃗

Direction of acceleration is down along the incline (+), direction of alpha=out of page
∑τ ⃗ =Iα ⃗

Forces parallel to the incline: +mg∙sin(theta)−fs=ma
Forces perpendicular to incline: +FN − mg∙cos=0
Torques about center of mass: fs∙r = I*alpha
Moment of Inertia I = ½ MR2 + (2/3)∙mr2
Rolling condition: a = r*alpha

Substitute and solve for a: a=(mg∙sinθ)/((I/r^2 +m))

Homework Equations

given above

The Attempt at a Solution

The part that i don't understand is how they are going from a = r*alpha to a=(mg∙sinθ)/((I/r^2 +m)), i am not seeing what they could be doing to get that answer. Any help is GREATLY appreciated, thanks.

 

[anathema]

I understand your confusion about the solution to this problem. Let me try to explain it step-by-step.

1. First, we need to understand the meaning of the equation a = r*alpha. This equation represents the relationship between the linear acceleration (a) and the angular acceleration (alpha) of a rolling object. In this case, the lander is rolling up the incline, so it is undergoing both linear and angular acceleration.

2. Next, we need to find the moment of inertia (I) of the lander. This is a measure of how difficult it is to change the rotational motion of the lander. For a uniform cylinder, the moment of inertia is given by I = ½ MR2, where M is the mass of the cylinder and R is its radius. We also need to consider the airbags, which can be thought of as a thin shelled sphere around the lander. The moment of inertia of a thin shelled sphere is given by I = (2/3)mr2, where m is the mass of the sphere and r is its radius.

3. Now, we can substitute the moment of inertia into the equation a = r*alpha. This gives us a = (r/I)*alpha. Since we have two different moments of inertia (one for the cylinder and one for the airbags), we need to add them together to get the total moment of inertia for the lander. This gives us a = (r/(½ MR2 + (2/3)mr2))*alpha.

4. Next, we need to consider the forces acting on the lander. There are two main forces to consider: the force of gravity (mg) and the force of friction (fs). We can break the force of gravity into its components parallel (mg*sin(theta)) and perpendicular (mg*cos(theta)) to the incline. The force of friction acts in the opposite direction of the motion, so it is negative (-fs).

5. Using the equation ∑F = ma, we can set up two equations: one for the forces parallel to the incline and one for the forces perpendicular to the incline. This gives us:

∑F_parallel = mg*sin(theta) - fs = ma
∑F_perpendicular = FN - mg*cos(theta) = 0 (since the lander is not moving up or down)

6. Finally, we can use the equation ∑τ