Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 6
A particle moves to two dimensional orbit defined by x(t) =A(2at-sinat) y(t)=A(1-cosat) A) Find the tangential acceleration a1 and normal acceleration an as function of time where the tangential and normal components are taken with respect to the velocity. B) Determine at what times in the orbit an has a maximum.
one year ago

Answers : (1)

Vikas TU
11137 Points
(1)  Tangential acceleration a1 = dv/dt = d^2x(t)/dt^2 
dx/dt = A(2a –acos(at))
dv/dt = A(a^2sin(at))
a1 = A(a^2sin(at))
Normal acceleration would be: d^2y(t)/dt^2 
dy/dt = Aasin(at)
d^2y/dt^2 = Aa^2cos(at) 
an = Aa^2cos(at) 
(2) For an to be maximum, d(an)/dt = 0
Aa^3sin(at) = 0
at = npi
t = n*pi/a it will be maximum. 
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details