Home
Class 11
PHYSICS
A ring rotates about z axis as shown in ...

A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration of a particle P (shown in figure) on the ring is `(6hat(i)-8hat(j)) m//s^(2)`. Find the angular acceleration of the ring & the abgular velocity at that instant. Radius of the ring is 2m.

Text Solution

Verified by Experts

The correct Answer is:
`-3hat(k) rad//s^(2), -2hat(k) rad//s`
`vec(a_(1))=6hat(i)=vec(alpha)xxvec(R)=vec(alpha)xx2hat(j)rArr vec(alpha)=-3hat(k) rad//s^(2)`
`vec(a_(r))=-8hat(j)=vec(omega)xxvec(v)=vec(omega)xx(omegaR)hat(i) rArr vec(omega)=-2hat(k) rad//s`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • KINEMATICS

    ALLEN|Exercise Exercise-04[B]|14 Videos

  • KINEMATICS

    ALLEN|Exercise Exercise-05 [A]|11 Videos

  • KINEMATICS

    ALLEN|Exercise Comprehension#7|3 Videos

  • ERROR AND MEASUREMENT

    ALLEN|Exercise Part-2(Exercise-2)(B)|22 Videos

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN|Exercise BEGINNER S BOX-7|8 Videos

Similar Questions

Explore conceptually related problems

A smooth ring of mass m and radius R = 1 m is pulled at P with a constant acceleration a= 4 ms^(-2) on a horizontal surface such that the plane of the ring lies on the surface. Find the angular acceleration of the ring at the given position. (in rad//s^(2) )

A particle is rotating about a fixed axis with angular acceleration vec alpha = (3hat i +hat j +hat k) rad/s^2 . The tangential acceleration of a point having radius vector ( 2 hat i-hat j -hat k) m from axis of rotation is (in m/s^2 )

Knowledge Check

  • A wheel is rolling on a horizontal plane. At a certain instant it has a velocity v and acceleration a of CM as shown in figure . Acceleration of

    A
    `A` is vertically upwards
    B
    `B` may be vertically down-wards
    C
    `C` cannot be horizontal
    D
    some point on the rim may he horizontal leftwards

  • A ring of radius R is rotating about axis of ring such that angular velocity is given as = 5t. Find acceleration of a point P on rim after 5 sec

    A
    5R
    B
    25R
    C
    `sqrt(650)`R
    D
    None of these

  • A ring of radius R rolls on a horizontal surface with constant acceleration a of the centre of mass as shown in figure. If omega is the instantaneous angular velocity of the ring. Then the net acceleration of the point of contact of the ring with gound is

    A
    zero
    B
    `omega^(2)R`
    C
    `a`
    D
    `sqrt(a^(2)+(omega^(2)R)^(2))`

    • Similar Questions

    Explore conceptually related problems

    Velocity of a particle at some instant is v=(3hat i + 4hat j + 5hat k) m//s . Find speed of the particle at this instant.

    The velocity and acceleration vectors of a particle undergoing circular motion are v = 2 hat(i) m//s and a = 2 hat(i)+4 hat(j)m//s^(2) respectively at some instant of time. The radius of the circle is

    At a particular instant velocity and acceleration of a particle are (-hat(i)+hat(j)+2hat(k))m//s and (3hat(i)-hat(j)+hat(k))m//s^(2) respectively at the given instant particle's speed is :

    A constant force F is applied at the top of a ring as shown in figure. Mass of the ring is M and radius is R. Angular momentum of particle about point of contact at time t

    A particle is moving in a circular path. The acceleration and momentum of the particle at a certain moment are a=(4 hat(i)+3hat(j))m//s^(2) and p=(8 hat(i)-6hat(j))"kg-m/s" . The motion of the particle is

    Home
    Profile