If a particle rotates along a circle of radius 3 m with an angular acceleration of `(pi)/(2) "rad"//s^(2)` starting from rest, then its average velocity over the time it covers quarter circle is :
A
`(pi)/(sqrt(2))m//s`
B
`(pi)/(2sqrt(2)) m//s`
C
`(2sqrt(2))/(pi)m//s`
D
`(pi)/(2) m//s`
Text Solution
Verified by Experts
The correct Answer is:
B
Let t be the time taken by the particle to go from A to B . ` :' theta = (1) /(2) alpha t^(2)` ` therefore (pi)/(2) =(1)/(2) xx(pi)/(4) xx t^(2) therefore t= 2s` ` V_(As) = ( "are" AB) /(t) =( r theta ) /(t) = ( sqrt(2 ) xxpi)/(2xx2) =(pi)/(2sqrt(2))m//s`
Topper's Solved these Questions
CIRCULAR MOTION
MARVEL PUBLICATION|Exercise TEST YOUR GRASP-1|1 Videos
CIRCULAR MOTION
MARVEL PUBLICATION|Exercise TEST YOUR GRASP-2|1 Videos
ATOMS, MOLECULES AND NUCLEI
MARVEL PUBLICATION|Exercise TEST YOUR GRASP|30 Videos
COMMUNICATION SYSTEMS
MARVEL PUBLICATION|Exercise TEST YOUR GRASP -20|10 Videos
Similar Questions
Explore conceptually related problems
Starting from rest, a particle rotates in a circle of radius R 2 m with an angular acceleration = /4 rad/s2.The magnitude of average velocity of the particle over the time it rotates quarter circle is:
A disc of radius 0.1 m starts from rest with an angular acceleration of 4.4 rad//s^(2) .Then linear velocity of the point on its after 5 s is
A particle is moving along a circle of radius 3 m .If its centripetal acceleration is 3ms^(-2) , its angular velocity in rad s^(-1) is
A particle starts from rest and moves with an angular acceleration of 3 rad //s^(2) in circle of radius 3 m . Its linear speed after 5 seconds will be
A particle moves along a circular path a radius 15 cm with a constant angular acceleration of 4 "rad/s"^(2). If the initial angular speed of the particle was 5 rad/s. Find its angular displacement in 5 seconds.
A particle starts moving at t = 0 in a circle of radius R = 2m with constant angular acceleration of a - 3 "rad/sec"^(2) . Initial angular speed of the particle is 1 rad/sec. At time t_(0) the angle between the acceleration vector and the velocity vector of the particle is 37^(@) . What is the value of t_(0) ?
A body rotates about a fixed axis with an angular acceleration of 3 rad//s^(2) The angle rotated by it during the time when its angular velocity increases frm 10 rad/s to 20 rad/s (in radian) is
A particle starts moving at t = 0 in a circle of radius R = 2m with constant angular acceleration of a - 3 "rad/sec"^(2) . Initial angular speed of the particle is 1 rad/sec. At time t_(0) the angle between the acceleration vector and the velocity vector of the particle is 37^(@) . What is the magnitude of the acceleration of the particle at t_(0) ?
A particle is revolving in a circular path of radius 2 m with constant angular speed 4 rad/s. The angular acceleration of particle is
MARVEL PUBLICATION-CIRCULAR MOTION-TEST YOUR GRASP-20