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Ostrovityanka [42]
3 years ago
6

A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot

in the vertical plane. The rod is held in the horizontal position and released from rest. What is the initial angular acceleration of the rod and the initial linear acceleration of the right end of the rod?
Physics
1 answer:
Leokris [45]3 years ago
5 0

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

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In this problem, we have:

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Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

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Explanation:

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