Question

A particle is moving with constant angular acceleration of 4 rad/s^2 in a circular path .At time T=0,PARTICLE was at rest . find the time at which the magnitude of centipetal acceleration and tangential acceleration are equal.

Answer 1

@Herp_Derp @henpen @hei @shadowfiend

Answer 2

@UnkleRhaukus

Answer 3

plz hlp me out

Answer 4

@Herp_Derp hlp me out

Answer 5

@Ruchi. I'm sorry I would've helped you out but honestly this is not my field.

Answer 6

i hav cprrect it

Answer 7

bt y c=0

Answer 8

Actually, I'm probably wrong-let me corret myself

Answer 9

As the particle starts at rest (at t=0, v=0, so only c=0 satisfies this)

Answer 10

it hav correct it and get t=1/2 now tell me y c=0,what is c here reffer to?

Answer 11

in think in case of differ. u hav refferd c.

Answer 12

\(\alpha\) is constant:\[a_c=\frac{v^2}{r}\]\[v=r\omega=r\alpha t\]\[a_c=\frac{(r\alpha t)^2}{r}=r\alpha^2t^2\]\[a_t=r\alpha\]\[a_t=a_c \Leftrightarrow r\alpha=r\alpha^2t^2\]\[\frac{1}{\alpha}=t^2\]\[t=\sqrt{\frac{1}{\alpha}}\]

Answer 13

henpen r u writng some thing or nt

Answer 14

Do you understand my work @Ruchi ?

Answer 15

no

Answer 16

Which part is confusing?

Answer 17

\[v=r \omega\] after this one

Answer 18

@Herp_Derp

Answer 19

dv/dt=4=angular acceleration
v=4t+c (c=0)
v=4t=angular velocity

tangential velocity=rw
=r4t

Centripetal acceleration=v^2/r
(r4t)^2/r=4

16r t^2=tangential acceleration=r(4)
t=sqrt(0.25)

Answer 20

\[\omega=\alpha t\]

Answer 21

Ruchi, c was just a possible initial condition, as there is no law that states that particles have to start at rest, so I was just generalising.

Answer 22

okay bt i hav'nt understood that rw=r4t

Answer 23

w=angularacceleration x t
angular acceleration here=4

Answer 24

r4t?

Answer 25

Sorry, tangential velocity=rw= r(4t)