Q: What is the centripetal acceleration of a point located 7.50 cm from the central axis of an ultracentrifuge spinning at a rate of 7.5 x 104 rev/min?
A: The given rate of 7.5 x 104 revolutions per minute is converted to radians per second ( ⍵ ) in the latter stages of the solution below.
First, determine what dimensions of a circle are used to define an angle. Draw a large circle ( C2 ) that surrounds a smaller circle ( C1 ). Next, draw a radius that begins at the center of ( C1 ) and terminates when it reaches the surface of ( C2 ). Label the radius contained by ( C1 ) as ( r1 ), and assume the longer line extending to ( C2 ) to be radius ( r2 ).
Next, draw an arc ( l1 ) along ( C1 ) that is equal in length to ( r1 ) and originates at the tip of ( r1 ). Likewise, draw an arc ( l2 ) along ( C2 ) that is equal in length to ( r2 ) and originates at the tip of ( r2 ). A length along the circumference of a circle that is equal in length to a circle’s radius is defined as a radian ( rad ).
Since ( C2 ) has a longer radius, a radian along ( C2 ) is longer than the corresponding radian along ( C1 ); HOWEVER, THE ANGLE BETWEEN THE BEGINNING AND END OF THE RADIANS ALONG ( r1 ) AND ( r2 ) ARE THE SAME. When the radian along either ( C1 ) or ( C2 ) is divided by its corresponding radius, the division yields a result equal in value. When dealing with any circle of any size, ( l / r ) = an angle theta ( θ ) radians. The total distance around any circle is ( 2𝛑r ), therefore, ( l / r ) for an entire circle yields [ ( 2𝛑r ) / ( r ) ] = ( 2𝛑 ). This is why it is stated that there are ( 2𝛑 rad ) in a complete circle. Since the full rotation of a circle is defined as having 3600 ( for convenience ), ( 2𝛑r ) = 3600. Likewise, there are ( 𝛑 ) radians ( 180 degrees ) in a half circle, and 1 rad = 57.30. Furthermore, angular speed is measured in radians per second ( ⍵ ).
( 7.5 x 104 rev / min )( 2𝛑 rad / 1 rev )( 1 min / 60 s ) = ( 4.7 x 105 rad / 60 s ) = ( 7.9 x 103 rad / sec ) = ( ⍵ ).
Since ( l / r ) = ( θ rad ), then ( l = ( r )( θ rad ) ). Dividing both sides by seconds gives us velocity ( v ) in meters per second on the left-hand side of the equation and ( r )( ⍵ ) on the right-hand side. Thus, the angular velocity ( v ) = ( r )( ⍵ ). Since ( v2 / r ) = α is the expression for angular centripetal acceleration, ( ac ), the following derivation is used to solve the problem:
( v ) = ( r )( ⍵ )
( v )2 = ( r )2( ⍵ )2 = ( r )( α )
( ac ) = ( v2 / r ) = ( r )( ⍵ )2 = ( α )
Note: In later discussions, we’ll see that torque ( 𝞃 ) = ( F )( r )( sin θ ). The value of torque ( twisting force ) is maximized when θ = 900. When this is the case, ( 𝞃 ) = ( F )( r ). Since ( F ) = ( m )( a ), ( 𝞃 ) = ( m )( a )( r ) = ( m )( r⍵2)( r ) = ( m )( r )( α ). Likewise, when both sides of ( v ) = ( r )( ⍵ ) are divided by seconds, ( ac ) = ( r )( α ).
Centripetal acceleration ( ac ) = ( 0.075 m )( 7.9 x 103 ⍵ )2. Finally, ( ac ) = ( 0.075 m )( 6.2 x 107 ⍵2 ) = ( 4.7 x 106 rad / s2 ).
RENAISSANCE PHYSICS 