Question

Magnetic tape is being fed over and around the light pulleys mounted in a computer. If the speed v of the tape is constant and if the magnitude of the acceleration of point A on the tape is 1.79 times that of point B, calculate the radius r of the smaller pulley.

Answer 1

The expression to obtain the radius of the smaller pulley is r = 0.559*R

If I have the value of R I will be able to calculate the value of r, but in the exercise you do not give it, so I leave the expression.

Step-by-step explanation:

The tangential components of acceleration at points A and B is equal to zero

`a_(A) =a_(B) =0`

The acceleration of points A and B are:

`a_(A) =(V^(2) )/(r) \\a_(B) =(V^(2) )/(R)`

Where

V = velocity of the tape

r = radius of the small pulley

R = radius of the large pulley

`(a_(A) )/(a_(B) ) =((V^(2) )/(r) )/((V^(2) )/(R) ) \\1.79=(V^(2)R)/(V^(2)r ) \\r=0.559*R`