- #1
saltine
- 89
- 0
Homework Statement
(a) Find the acceleration at the tip of the length L rod
(b) Find the normal force at the hinge
Homework Equations
Moment of Inertia of rod: I = mL2/3
The Attempt at a Solution
(a)
Since the rod is rotating, there is centripetal acceleration throughout the rod. This points toward the hinge.
acen = - L ω2 er ... (1)
The fact that the rod is attached to the hinge would cancel all translational forces.
The only remaining force comes from the net torque. The net torque at the hinge is:
τnet = - τ + (L/2) mg sinθ ... (2)
The net torque is related to the angular acceleration α by the moment of inertia. And α is related to the tangential a by the rod length.
τnet = I α
atan = L α
atan = L τnet/I = (3/mL)τnet eθ ... (3)
This is counter-clockwise.
The total a is acen + atan ... (4)
(b)
The only force is mg, so the normal force at the hinge is mg (pointing up).
This seems right to me, what do you think?
Now there is a weird problem:
Suppose I also calculate the net force at the tip. Since the translational forces are cancelled, the net force at the tip comes purely from the net torque:
Ftip = τnet/L ... (5)
This force is entirely tangential. Since I also had the acceleration at the tip, I could compute the "mass" at the tip:
Ftip = mtip atip ... (6)
Now, this is completely bizzare, since F and a do not point in the same direction, and m is supposed to be 0 at the tip. What is the logical reason that (6) is not valid? (Or perhaps (5) is not valid (?))
- Thanks