Lets do
We know
The rate of change of velocity is acceleration .
Integrate both sides
As acceleration is constant .Take it outside of integral .On velocity we can take limit u to v and time from 0 to t
Hence
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Answer: a) angular acceleration(alpha)=-14.8rad/s^2
b) time taken (t) = 1.52s
Explanation:
What we are given
Mass of the solid sphere m =225g = 0.225kg
Diameter D = 3.00cm = 0.0300m
Radius = D/2 = 0.01500m
Frictional Force = 0.0200N
a) to determine the angular acceleration, we first calculate the torque, then moment of inertia, before the angular acceleration.
Torque = -fr
= - (0.0200)(0.01500)
=-3.00X10^-4Nm
Moment of inertia I
= 2/5mr^2
=2/5(0.225)(0.01500)^2
=2.025X10^-5kgm^2
Angular acceleration (alpha)= torque/moment of inertia (I)
= -3.00X10^-4Nm/2.025X10^-5kgm^2
=-14.8rad/s^2
b) time taken (∆t) = w/alpha
w= -22.5rad/s
Angular acceleration (alpha) = -14.8rad/s^2
∆t = -22.5/-14.8
= -1.52s
Answer
The second rock will land 2.4s after the first rock
Explanation:
Given that
Height of the building s=50m
We assume that the first rock is acting with gravity so that a=9.81m/s
And initial velocity u=0
Applying the equation of motion
S=ut+1/2at²
50=0*t+1/2(9.81)t²
50=4.905t²
t²=50/4.905
t²=10.19
t=√10.19
t=3.19sec
For the second rock initial velocity u=8m/s and v=0 and a=9.81
Applying the equation of motion
v=u+at
0=8+9.81t
t=-8/9.81
t=0.81sec
Hence the second rock will land 2.4s after the first rock
I.e
3.19-0.81
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