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

Answer:

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

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

( Acceleration = ( About ) 9290 m / [tex]s^2[/tex]

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 ) [tex]a_{rad}[/tex] = w^2( r ),

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

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

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

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

( Acceleration = ( About ) 9290 m / [tex]s^2[/tex]

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 ) [tex]a_{rad}[/tex] = w^2( r ),

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

[tex]a_{rad}[/tex]

= ( About ) 9290 m /[tex]s^2[/tex]