When a disc rotates so as to vary its angular velocity with time as ω= 2t - 3t^{2}[/tex] .
find the total angular displacement of any point on disc relative to its centre when the disc will stop momentarily.
Answers
Answer:
A disc rotates about a fixed axis. Its angular velocity ω varies with time according to the equation ω=at + b. Initially at t = 0 its angular velocity is 1.0 rad/s and angular position is 2 rad; at the instant t= 2s angular velocity is 5.0 rad/s. Determine angular position θ and angular acceleration when t = 4s.
MEDIUM
ANSWER
We have
w=at+b
at t=Os,w=1 rod/s
So, 1=O+b
At t=2 & w=5 rod/s
so, 5=2a+1
⇒ a=2 rod/s
2
w=2t+1
α=
dt
dw
=
dt
d(2t+1)
=2 rad/s
w=
dt
dθ
=(2t+1)dt
⇒ dθ=(2f+1)dt
⇒ ∫dθ=∫(2f+1)dt
⇒ [θ]
θ
0
θ
=[t
2
+t]
0
t
⇒ θ−2=t
2
+t
⇒ θ=t
2
+t+2
`8 rad//s^(2)`
`10 rad//s^(2)`
`12 rad//s^(2)`
None of these
Answer :
C
Solution :
Given `omrga=theta^(2)+2theta`
`(domega)/(dtheta)=2theta+2`
`alpha=omega(domega)/(dtheta)=(theta^(2)+2theta)(2theta+2)`
at `theta=1`
`alpha=12 rad//sec^(2)`