A bar is constrained to rotate on a horizontal surface around an axis perpendicu

Question

A bar is constrained to rotate on a horizontal surface around an axis perpendicular to the rod and passing through the dot. a) What is the net torque on the bar at the instant shown about the axis indicated by the dot? Write the answer to three significant figures. Remember - torque is a vector! (2.46 N m, ccw) b) If the moment of inertia of the rod is 0.182 kg m^2. what is the angular acceleration of the rod around the axis? Write the answer to three significant figures. {13.5 rod/s, ccw) c) Assume the net torque remains constant. Starting from rest, how many revolutions will the bar make In 2.50 s? {6.73 rev}

Explanation / Answer

Here ,

part a) at this moment , about the pivot point

net torque on the bar = 10 * sin(30) * 0.75 - 8 * sin(40) * 0.25

net torque on the bar = 2.46 N.m in counter clockwise direction

b<)

as the moment of inertia = 0.182 Kg.m^2

Using the second equation of motion

I * angular acceleration = torque

0.182 * angular acceleration = 2.46

solving

angular acceleration = 13.5 rad/s^2

direction will be same as the torque , direction of angular acceleration is counter clockwise

b)

for the angle rotated in 2.50 s

angle rotated = 0.50 * 13.5 * 2.50^2

angle rotated = 42.1 radian

angle rotated = 42.1/(2pi) revs

angle rotated = 6.73 revs

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