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sattari [20]
3 years ago
12

A thin, horizontal rod with length l and mass M pivots about a vertical axis at one end. A force with constant magnitude F is ap

plied to the other end, causing the rod to rotate in a horizontal plane. The force is maintained perpendicular to the rod and to the axis of rotation.
Calculate the magnitude of the angular acceleration of the rod.
Physics
1 answer:
katen-ka-za [31]3 years ago
8 0

Answer:

The magnitude of the angular acceleration is α = (3 * F)/(M * L)

Explanation:

using the equation of torque to the bar on the pivot, we have to:

τ = I * α, where

I = moment of inertia

α = angular acceleration

τ = torque

The moment of inertia is equal to:

I = (M * L^2)/3

Also torque is equal to:

τ = F * L

Replacing:

I * α = F * L

α = (F * L)/I = (F * L)/((M * L^2)/3) = (3 * F)/(M * L)

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Answer:

<h2>Mass of 1 Kg and 2 Kg, 1 meters apart.</h2>

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Answer:

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4 0
2 years ago
A 0.5 kg stone is raised from 1m to 2m height from the ground. what is the change in potential energy of the stone?
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Given: The mass of stone (m) = 0.5 kg

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Acceleration due to gravity (g) = 9.8 m/s²

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where, all alphabets are in their usual meanings.

Now, we shall calculate the change in potential energy of the stone

Δ P = P₂ - P₁ = mg (h₂ - h₁)

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Answer:

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