Calculating Gate's Torque, KE & Angle When Bull Rushes Out

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The discussion focuses on calculating the torque, rotational kinetic energy, angle of rotation, and whether a gate will close when a bull rushes toward it. The torque applied by the farmer is calculated as 300 Nm. The rotational kinetic energy after 3 seconds is determined to be 405 J, and the gate rotates 1.35 radians in that time. Since 1.35 radians is less than the required π/2 radians to close the gate, it will not be closed in time. The calculations emphasize the importance of using proper units for energy and understanding angular motion principles.
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Homework Statement


A gate to a bullpen is open at a right angle to the fence and the bull is rushing toward the opening to get out. The farmer estimates that the bull will reach the opening in 3 sec. so he pushes at the end of the gate (always at a right angle) with a force of 100N. The moment of inertia of the gate about the hinge is 1,000 kg m^2 and the gate length is 3.

a. What is the magnitude of the torque applied by the farmer.
b. What is the rotational kinetic energy of the gate after 3 seconds
c. Through what angle will the gate have rotated after 3 sec.
d. Will the gate be closed.

Homework Equations



Torque, KE roational, Angular Kinematics

The Attempt at a Solution



a. Torque= rFsin(theata) = (3)(100)sin90= 300Nm
b. KE= (1/2)(I)(w^2) = (1/2)(1000)(?)
c. angle= at^2/2?
d. ?
 
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Part a is correct; for parts b, c, and d, you first need the equation (derived from Newton's 2nd law) that allows you to calculate the angular acceleration caused by a net torque.
 
If I use torque=Ia and solve for a I will get a=torque/I = (300)/(1000) = .3 rad/sec^2 as the angular acceleration. For the w value after 3 sec it would be .9rad/sec? So I could then do the KE= (1/2)(I)(w^2) = (1/2)(1000)(.9^2) = 405 Then for part c could I use Theta= at^2/2 and get (.3)(3^2)/2 = 1.35 radians. Is any of this right? What would I do for the last part?
 
tachu101 said:
If I use torque=Ia and solve for a I will get a=torque/I = (300)/(1000) = .3 rad/sec^2 as the angular acceleration. For the w value after 3 sec it would be .9rad/sec? So I could then do the KE= (1/2)(I)(w^2) = (1/2)(1000)(.9^2) = 405 Then for part c could I use Theta= at^2/2 and get (.3)(3^2)/2 = 1.35 radians. Is any of this right? What would I do for the last part?
Yes, excellent. Don't forget the units for energy, 405 __? For the last part, you know that the gate has rotated 1.35 radians. How many radians would it need to rotate to a completely closed position, given that it started out at a right angle to the fence?
 
pi/2 would be the angle needed to close the gate, but 1.35 is less than pi/2, so it would not be closed in time.
 
tachu101 said:
pi/2 would be the angle needed to close the gate, but 1.35 is less than pi/2, so it would not be closed in time.
Correct. Now don't forget the units for the kinetic energy.
 
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