Problem 20 You are holding a block at rest on top of a horizontal frictionless s

Question

Problem 20 You are holding a block at rest on top of a horizontal frictionless surface. The block has a mass 1.91 kg and is tied to a string that stays horizontal until it goes over the outer rim of a pulley, which has a diameter of 0.200 m , and connects to a block of mass 3.10 kg that is suspended straight vertically from the other end of the string. When you release the system from rest, you observe that both the blocks (i.e. each block) move a distance of 1.20 m over a time interval of 0.750 s . You are not told whether the pulley is in the shape of a wheel with spokes or in the shape of a uniform disk, i.e. you are not told how the mass is distributed within the structure of the pulley. All you know is that the pulley is not massless and has a radius (has circular geometry) as given above and the pulley is rotating about its center which is also its center of mass. Part A From this information, is it possible to calculate the moment of inertia of the pulley about its axis of rotation? If yes, calculate it and enter your answer correct to FIVE sig figs. If no, enter 0.0000 Take the free fall acceleration to be g = 9.80 m/s^2 .

Explanation / Answer

A) yes.

Let I is the moment of inrtia of pulley.

radius of pulley, r = d/2 = 0.2/1 = 0.1 m

Let a is the acceleration of each block.

Apply,

s = u*t + 0.5*a*t^2

s = 0.5*a*t^2 (since u = 0)

a = 2*s/t^2

= 2*1.2/0.75^2

= 4.27 m/s^2

let m1 = 1.91 kg

m2 = 3.1 kg

tension in the string connected to m1, T1 = m1*a

= 1.91*4.27

= 8.16 N

tension in the string connected to m2, T2 = m2*g - m2*a

= 3.1*9.8 - 3.1*4.27

= 17.14 N

now Apply net torque acting on pulley = I*alfa

T2*r - T1*r = I*a/r

I = (T2 - T1)*r^2/a

= (17.14 - 8.16)*0.1^2/4.27

= 0.021 kg.m^2 <<<<<<<<-------------Answer

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