Rotational Motion: Disc Movement

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SUMMARY

The discussion focuses on the analysis of a grinding wheel's rotational motion, specifically a 750-gram solid disc with a diameter of 25.0 cm, operating at a constant speed of 220 rpm (23.04 rad/s). Key calculations include the rotational inertia, which is determined to be 5.86 x 10^-2 kgm², and the angular acceleration, calculated as 0.512 rad/s². The user seeks clarification on the initial angular velocity and the torque exerted by friction, indicating a need for precise understanding of angular motion equations.

PREREQUISITES
  • Understanding of rotational inertia and its calculation (Idisc = 0.5(mr²))
  • Knowledge of angular velocity conversion from rpm to rad/s
  • Familiarity with angular acceleration and its relationship to torque (α = T/I)
  • Basic principles of kinematics in rotational motion
NEXT STEPS
  • Learn about the relationship between linear and angular motion (v = rω)
  • Study the equations of motion for rotational dynamics
  • Explore torque calculations and their applications in rotational systems
  • Investigate the effects of friction on angular deceleration
USEFUL FOR

Students studying physics, particularly those focusing on rotational dynamics, engineers working with mechanical systems, and anyone interested in the principles of angular motion and torque calculations.

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Homework Statement


A 750 gram grinding wheel 25.0 cm in diameter is in the shape of a uniform solid disc. When it is in use it turns at a constant 220 rpm about an axis perpendicular to its face through its center. When the power is switched off the wheel stops in 45.0 s with constant angular acceleration due to friction

A) What is rotational intertia of the wheel?
B) What is the initial angular velocity in rad/s?
C) What is the constant angular acceleration in rad/s^2?
D) What is the magnitude of the torque exerted by friction in N-m?

m = 0.750 kg
r= 0.125 m
wf = 23.04 rad/s (converting 220rpm)
t= 45.0 s

Homework Equations


Part A) Idisc= 0.5(mr^2) = 0.5(0.750kg)(0.125m^2) = 5.86*10^-2 kgm^2

The Attempt at a Solution



I was thinking along these lines... I could find part C first (angular acceleration - alpha) and use that to find part B (initial angular velocity - w0) by using this equation : wf = w0 + alpha(time). I could then find part D (torque - T) by using this equation : alpha = T/I

My problem is finding angular acceleration, or alpha.

This is what I did, please let me know if it's correct or not.

v(speed)= radius*wf
v = (0.125m)(23.04 rad/s) = 2.88
vf= v0 +a(linear)(time)
2.88 = 0 +a(45seconds)
a(linear) = 0.064 m/s
a(linear) = radius*a(angular)
0.064 = 0.125*a
alpha = 0.512 rad/s^2

If it's wrong, could you please advise on how to get the right answer?
Thank you in advance :)Disclaimers: This is my first post, so please be gentle! Also, this is not homework, it is a practice test. Thanks again!
 
Last edited:
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Did I read this wrong... I assumed wf= 220rpm = 23.04 rad/s. Is the 220rpm actually w0? And if so, am I just converting it wrong?
 

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