N
nikster
Guest
A particle P moves in a plane with constant angular velocity 'w' about O. If the rate of growth of acceleration is wholly radial, prove that:
r'' = 1/3 rw^2
If initially the particle is at A and starts with velocity 'aw' perpendicular to OA, show that at time t,
(r')^2 = 1/3 (w^2 - a^2)
here r'' -- second derivative of r with respect to t
r'-- first derivative of r with respect to t
r- radius and r is a function of t
any help will be appreciated! any clues, maybe if you get the first/second part. thanks in advance~!!
maybe you have to use radial and transverse components of acceleration to solve this..
radial acceleration = r'' + rw^2
transverse component = rw' + 2r'w
r'' = 1/3 rw^2
If initially the particle is at A and starts with velocity 'aw' perpendicular to OA, show that at time t,
(r')^2 = 1/3 (w^2 - a^2)
here r'' -- second derivative of r with respect to t
r'-- first derivative of r with respect to t
r- radius and r is a function of t
any help will be appreciated! any clues, maybe if you get the first/second part. thanks in advance~!!
maybe you have to use radial and transverse components of acceleration to solve this..
radial acceleration = r'' + rw^2
transverse component = rw' + 2r'w