A particle of mass m is moving along a trajectory given by x= x0+ a cos
, y = y0 + b sin
t The torque, acting on the particle about the origin, at t=0 is :
(a)
−m(xob
)
(b)
m(−xob + yoa)
)
(c)
+myoa
(d)
Zero
A particle of mass m is moving along a trajectory given by x= x0+ a cos , y = y0 + b sin t The torque, acting on the particle about the origin, at t=0 is :
(a)
−m(xob )
(b)
m(−xob + yoa) )
(c)
+myoa
(d)
Zero
Question ID - 51322 | SaraNextGen Top Answer
A particle of mass m is moving along a trajectory given by x= x0+ a cos , y = y0 + b sin t The torque, acting on the particle about the origin, at t=0 is :
(a)
−m(xob )
(b)
m(−xob + yoa) )
(c)
+myoa
(d)
Zero
|
A particle of mass m is moving along a trajectory given by x= x0+ a cos |
|||
|
(a) |
−m(xob |
(b) |
m(−xob + yoa) |
|
(c) |
+myoa |
(d) |
Zero |
1 Answer
127 votes
Answer Key / Explanation : (c) -
x= x0+ a cos
y = y0 + b sin
t
⇒ vx = −a
sin(
t), vy =b
cos(
t),
ax =−a
cos(
t), ay =−b
sin(
t),
At t=0, x=x0 +a, y=y0
ax =−a
, ay =0
=m(−a
)x y0(−
)
=+my0a
.
127 votes
Answer Link
127