Fewpal
0 votes
27 views
in Physics by (58.6k points)

A straight link AB 40cm long has, at a given instant, its end B moving along line OX at 0.8m/ s and acceleration at 4m/sec2 and the other end A moving along OY, as shown in fig. Find the velocity and acceleration of the end A and of mid point C of the link when inclined at 300 with OX.

1 Answer

+1 vote
by (66.3k points)
selected by
 
Best answer

Let the length of link is L = 40cm and AD = Y, OB = X

X2 + Y2 = 1 ...(i)

Diff with respect to time, and -ive sign is taken for down word motion of A, when B is moving in +ive direction, we get

2Xdx/dt – 2Ydy/dt = 0XVB – YVA = 0

VA = (X/Y)VB = (Lcosθ/Lsinθ)VB = VB/tanθVA 

= 0.8/tan300 = 1.38m/sec ...(iii)

VA = 1.38m/sec

Again differentiating equation (2), we get

Xd2x/dt2 + (dx/dt)2 – yd2y/dt2 – (dy/dt)2 = 0

X.aB + (VB)2 – Y.aA – (VA)2 = 0

0.4cos 300 X 0.4 – (0.8)2 – 0.4sin 300 X aA – (1.38)2= 0

1.38 + 0.64 – 0.2aA – 1.9 = 0

aA = 0.6.6m/sec2

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...