What force must be applied to the end of a rod along the x-axis of length 2.25 m in order to produce a torque on the rod about t
1 answer:
Answer:
<h2>4.9N approx.</h2>Explanation:
Step one:
given data
length of rod r= 2.25m
Required torque = 11 Nm
Required
The force needed to produce a torque of 11Nm
Step two:
<em>"By definition, Torque is the twisting force that tends to cause rotation. </em>
<em>The point where the object rotates is known as the axis of rotation."</em>
Mathematically,
= torque
r = radius
F = force
= angle between F and the lever arm
in this case is zero
substituting we have
F=11/2.25
F=4.88N
F=4.9N approx.
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Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,
Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀ =A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values
b)
, where
is time period of damped SHM
⇒
let be number of oscillations made
then,
⇒