a wheel 0.35m in diameter rotates at 2200rpm. calculate its angular velocity in rad/s and its linear speed and acceleration of a point on th

Question

a wheel 0.35m in diameter rotates at 2200rpm. calculate its angular velocity in rad/s and its linear speed and acceleration of a point on the edge of the wheel. need help asap

in progress 0
Thiên Thanh 4 years 2021-08-16T10:33:34+00:00 1 Answers 8 views 0

Answers ( )

  1. kimcuc
    0
    2021-08-16T10:35:28+00:00
    Reply

    Answer:

    ( Angular Velocity = ( About ) 230 rad / s,

    ( Linear Speed = ( About ) 40.25 m / s,

    ( Acceleration = ( About ) 9290 m / s^2

    Explanation:

    Here we want the angular velocity in radians per second, the linear velocity and acceleration.

    The diameter = .35 meters, and thus we can conclude that the radius be half of that, or .175 meters. For part ( a ), or the calculation of the angular velocity, it is given that the diameter rotates at 2200 revolutions per minute – but we need to convert this into radians per second.

    We can say that there are 2π radians for every minute, and for every minute there are 60 seconds. Therefore –

    ( a ) w = 2,200 rpm( 2π rads / rev )( 1 min / 60 sec )…

    Hence, w = ( About ) 230 rad / s

    _____

    For this second part we can calculate the the linear velocity by multiplying the angular velocity ( omega ) by the radius r –

    ( b ) v = w( r ) – Substitute,

    v = ( 230 rad / sec )( .175 m )…

    v = ( About ) 40.25 m / s

    _____

    And for this last bit here, to find the acceleration we can simply take the angular velocity ( omega ) squared, by the radius r –

    ( c ) a_{rad} = w^2( r ),

    a_{rad} = ( ( 230 rad / sec )^2 )( .175 m )…

    a_{rad} = ( About ) 9290 m / s^2

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )