7 please help this is my 3rd time posting this A yo-yo has a rotational inertia

Question

7 please help this is my 3rd time posting this

A yo-yo has a rotational inertia of 1080 g.cm2 and a mass of 120 g. Its axle radius is 3.8 mm, and its string is 80 cm long. The yo-yo rolls from rest down to the end of the string. What is the magnitude of its linear acceleration? How long does it take to reach the end of the string? As it reaches the end of the string, what is its linear speed? As it reaches the end of the string, what is its translational kinetic energy? As it reaches the end of the string, what is its rotational kinetic energy? As it reaches the end of the string, what is its angular speed?

Explanation / Answer

(a)
Consider translational motion. The Yo-Yo is dragged down by gravity while an unknown tension T in the string acts in opposite direction: Hence:
m · a = m · g - T
<=>
T = m·(g - a)
The rotatory motion is accelerated by the torque exerted by the tension T on the axis of radius r:
J · = T·r
<=>
T = J·/r
Equate the two equations to eliminate T
m·(g - a) = J·/r
As long the string does not slips at the axis, rotatory motion and angular motion are related as:
s = ·r => v = ds/dt = r·d/dt = r· => a = dv/dt = r·d/dt = ·r
Hence:
m·(g - a) = J·a/r²
=>
a = g / (1 + J/(m·r²))
= 9.81m/s² / (1 + 1080gcm² / ( 120g · (0.38cm)²))
= 0.157m/s²

(b)
The displacement of a constantly accelerated motion starting form rest is
s = a·t²/2
=>
t = (2·s/a)
= ( 2 · 8m / 0.157m/s²)
= 3.19s

(c)
The instantaneous velocity in a constantly accelerated motion starting form rest is
v = a·t
at t = (2·s/a)
v = a·(2·s/a) = (2·s·a)
= ( 2 · 0.8m · 0.157m/s² )
= 0.501m/s

(d)
E_kin,trans = m·v²/2
for v = (2·s·a)
E_kin,trans = m·s·a
= 0.120kg · 0.8m · 0.157m/s²
= 0.015J

(e)
E_kin,rot = J·²/2
As shown in a = v/r, thus:
E_kin,rot = J·(v/r)²/2
for v = (2·s·a)
E_kin,rot = J·s·a / r²
= 1.08kgcm² · 0.8m · 0.157m/s² / (0.38cm)²
= 0.939J

(f)
= v/r
= 0.501m/s / 0.0038m
= 131.84 s^-1

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