How to Calculate Time, Speed, and Angular Speed for Rotational Dynamics?

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SUMMARY

The discussion focuses on calculating time, speed, and angular speed for a hoop with a radius of 0.0690 m and a mass of 0.152 kg. The acceleration of the hoop's center of mass is determined to be 4.9 m/s², and the tension in the string while the hoop descends is 0.745 N. Using kinematic equations, the time taken to descend 0.600 m is calculated as 0.495 s, the speed of the hoop's center of mass after descending is 2.426 m/s, and the angular speed is 35.152 rad/s. These calculations utilize fundamental principles of rotational dynamics and kinematics.

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squintyeyes
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how do you do part c, d, and e

A string is wrapped several times around the rim of a small hoop of radius 0.0690 m and a mass 0.152 kg. The situation is sketched in the figure below.

If the free end of the string is held in place and the hoop is released from rest, calculate the following quantities:

(a) the magnitude of the acceleration of the hoop's center of mass
4.9 m/s2

(b) the tension in the string while the hoop descends and the string unwinds
0.745 N

(c) the time it takes the hoop to descend 0.600 m
_________ s

(d) the speed of the hoop's center of mass after it's descended 0.600 m
__________m/s

(e) the angular speed of the rotating hoop after it's descended 0.600 m
__________rad/s


Attempt
Σ Forces in x direction = ma
mg - Tension = ma
mg - T= ma

Σ Torque = Iα
Iα = Tr
(mr^2)(a/r)= Tr
T= ma

mg - T = ma
mg- ma = ma
mg = 2ma
g = 2a
a = g/2 = 9.8/2= 4.9

T = ma
= (0.152)(4.9)
= 0.7448
 

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squintyeyes said:
how do you do part c, d, and e
You found the acceleration; use kinematics to solve those parts of the problem.
 
so for part c, would i just do

Δy = vt + (1/2)at^2
0.6 = 0 + (1/2) (4.9) t^2
t^2 = 0.244897
t = 0.495 s

then d is found by

final velocity = inital velocity + at
= 0 + (4.9)(0.495)
= 2.426 m/s

Lastly e is found by

σ = v/r
= 2.426/0.069
= 35.152 rad/s

thanks i already put entered them and they are correct!
 
Last edited:

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