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

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Homework Help Overview

The problem involves calculating the torque, rotational kinetic energy, and angle of rotation of a gate being pushed open by a farmer as a bull rushes toward it. The context is rooted in rotational dynamics, specifically focusing on torque and angular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of torque and its application to find angular acceleration. There are attempts to derive the rotational kinetic energy and angle of rotation using relevant equations. Some participants question the correctness of their calculations and seek clarification on the final outcome regarding the gate's position.

Discussion Status

Some participants have provided calculations for torque and angular acceleration, while others have confirmed the correctness of these calculations. There is ongoing exploration of the implications of the calculated angle of rotation in relation to whether the gate will close.

Contextual Notes

Participants are working under the constraints of the problem statement, including the given force, moment of inertia, and time frame for the bull's approach. There is a focus on ensuring proper units are used in calculations, particularly for energy.

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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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