A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad/`s^(2)`. Its net acceleration in m/`s^(2)` at the end of 2.0 s is approximately
A
`7.0`
B
`6.0`
C
`3.0`
D
`8.0`
Text Solution
Verified by Experts
Angular acceleration, `alpha = 2 "rad"//s^(2)` diameter of the disc, R = 50 cm = 0.5 m Angular speed of the disc after 2 s, `omega = omega_(0) + alpha` t = `(0 + 2 xx 2) " "[ because omega_(0) = 0]` At that instant, radial acceleration, `a_(r ) = R omega^(2) = 0.5 xx (4)^(2) = 8 m//s^(2)` and tangential acceleration, `a_(t) = R alpha = 0.5 xx 2 = 1 m//s^(2)` `therefore` Resultant acceleration, a = `sqrt(a_(r )^(2) + a_(t)^(2)) = sqrt( 8^(2) + 1^(2)) approx 8 m//s^(2)` The option D is correct.
Calculate the radius of gyration of a uniform thin rod of mass M and length L about an axis of rotation perpendicular to its length and passing through its centre.
What is the force which produces an acceleration of 1 m//s^(2) in a body of mass 1 kg ?
A circular ring of diameter 40 cm and mass 1 kg is rotating about an axis normal to its plane and passing through the centre with a frequency of 10 rps. Calculate the angular momentum about its axis of rotation.
A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is would round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s^(-2) is
A lift of mass 400 kg is hung by a wire. Calculate the tension in the wire when the lift is (a) at rest, (b) moving upward with a constant acceleration of 2.0 m//s^(2) and (c) moving downward with an acceleration 2.0 m//s^(2) .
CHHAYA PUBLICATION-CIRCULAR MOTION-EXAMINATION ARCHIVE WITH SOLUTIONS (NEET)
