Answer:
The time after starting is 0.178 sec.
Explanation:
Given that,
Rotational speed = 600 rev/min
Time t = 2 sec
Angle = 45
Distance OP= 6''
According to figure,
We need to calculate the angular acceleration
Using formula of angular acceleration
[tex]\alpha=\dfrac{\omega}{t}[/tex]
[tex]\alpha=\dfrac{\dfrac{2\piN}{60}}{t}[/tex]
Put the value into the formula
[tex]\alpha=\dfrac{\dfrac{2\times\pi\times600}{60}}{2}[/tex]
[tex]\alpha=10\pi\ rad/s62[/tex]
We need to calculate the tangential acceleration
Using formula of tangential acceleration
[tex]a_{t}=OP\times \alpha[/tex]
Put the value into the formula
[tex]a_{t}=6\times10\pi[/tex]
[tex]a_{t}=60\pi\ in/s^2[/tex]
We need to calculate the normal acceleration
Using formula of normal acceleration
[tex]a_{n}=OP\times\omega^2[/tex]
For angle [tex]\theta=45[/tex]
[tex]a_{n}=a_{t}[/tex]
Put the value into the formula
[tex]OP\times\omega^2=60\pi[/tex]
[tex]6\times\omega^2=60\pi[/tex]
[tex]\omega^2=10\pi[/tex]
[tex]\omega=\sqrt{10\pi}[/tex]
[tex]\omega=5.605\ rad/s[/tex]
We need to calculate the time
Using the kinematics equation
[tex]\omega=\omega_{0}+\alpha t[/tex]
Put the value into the formula
[tex]5.605=0+10\pi t[/tex]
[tex]t=\dfrac{5.605}{10\pi}[/tex]
[tex]t=0.178\ sec[/tex]
Hence, The time after starting is 0.178 sec.
