practice problem 1
|Page 1: The questions||Page 2: The answers|
The answers are also available on the second page of the worksheet.
practice problem 2
- the moment of inertia coefficient by timing the trip from top to bottom
- the critical angle past which an object will slip rather than roll down the incline
This problem is best solved using the conservation of energy. The rolling body starts with gravitational potential energy at the top of the ramp and ends with translational and rotational kinetic energy at the bottom. Since the object isn't slipping, it's angular velocity is v/R.
Ug = Kt + Kr Mgh = 1 Mv2 + 1 Iω2 2 2 Mgh = 1 Mv2 + 1 I ⎛
2 2 R 2MghR2 = MR2v2 + Iv2 I = 2MghR2 − MR2v2 v2 I = ⎛
2gh − 1 ⎞
Here's the coefficient. It still needs a bit of work.
α = 2gh − 1 v2
Recall that for an object accelerating uniformly from rest, its final speed is twice its average speed.
v = Δs = v + v0 = v + 0 Δt 2 2 v = 2Δs Δt
Substitute this expression into the formula for the coefficient. (We'll drop the ∆ [delta].)
α = 2gh − 1 v2 α = 2gh − 1 (2s/t)2 α = ght2 − 1 2s2
If you prefer to measure the angle of inclination of the ramp rather than its height you get a slightly different formula.
α = ght2 − 1 2s2 α = g(s sin θ)t2 − 1 2s2 α = gt2sin θ − 1 2s
The second question is best solved using Newton's laws of motion. The component of the weight parallel to the incline pulls the object down the incline while the frictional force pulls it up. Friction also exerts a torque that makes the object rotate about its center of mass.
translational ∑F = ma W∥ − f = Ma∥ Mg sin θ − μMg cos θ = Ma∥ g sin θ − μg cos θ = a∥
Pay special attention the α [alpha] symbols. Sometimes α means angular acceleration and sometimes α is the coefficient of the moment of inertia. (The switch takes place between the second and third lines in the work shown below.)
rotational ∑τ = Iα W∥R = Iα μMg(cos θ)R = (αMR2)(a∥/R) μg cos θ = αa∥
Divide these two equations to eliminate the acceleration parallel to the ramp and solve for the critical angle (or its tangent).
g sin θ − μg cos θ = a∥ μg cos θ αa∥ tan θ − 1 = 1 μ α
As the angle increases, friction decreases. Eventually the static friction force won't be strong enough to spin the object and it will slip. The critical angle at which this transition takes place is…
tan θ = μ ⎛
1 + 1 ⎞
This formula is similar to one that was derived in an earlier part of this book (tan θ = μ). That formula was for an object on an incline that doesn't slip or roll.
practice problem 3
practice problem 4