4/5/2016 11:00 PM A 25/100 4/5/2016 08:13 PM Gradebook Print Calculator Periodic

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4/5/2016 11:00 PM A 25/100 4/5/2016 08:13 PM Gradebook Print Calculator Periodic Table Question 8 of 16 Map pling A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are 0.703 m and 1.27 kg. What constant-magnitude force acting at the other end of the rod perpendicularly both to the rod and to the axis will accelerate the rod from rest to an angular speed of6.37 rad/s in 8.47 s? Number O Previous ® Give Up & View Solution Check Answer Next H Ext Hint

Explanation / Answer

here

the torque on the rod is T = I * alpha

I = moment of inertia

alpha = angular acceleration

I = m * L^2 / 3

I = 1.27 * 0.703^2 / 3 = 0.2092 kg m^2

then by using first equation of motion

w = w0 + alpha * t

6.37 = 0 + alpha * 8.47

alpha = 0.752 rad/s^2

put values in the formula of torque

T = 0.2092 * 0.752 = 0.157 Nm

then

T = F * L

F = 0.157 / 0.703 = 0.223 N

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