Question

A uniform disk (I =½ MR2 ) of mass 8.0 kg can rotate without friction on a fixed axis. A string is wrapped around its circumference and is attached to a 6.0 kg mass. The string does not slip. What is the tension in the cord while the mass is falling?

Answer 1

T = 23.54 N

Step-by-step explanation:

Given,

• mass of the pulley = M = 8.0 kg
• mass of the object = m= 6.0 kg
• moment of inertia of the disk = I =
  `(1)/(2)MR^2`

Let 'T' be the tension in the string and 'a' be the acceleration of the block,

Hence, the angular acceleration of the pulley =
`\alpha\ =\ (a)/(R)`

From the f.b.d. of the mass

`mg\ -\ T\ =\ ma\\\Rightarrow a\ =\ (mg\ -\ T)/(m)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1),`

From the f.b.d. of the pulley,

`\sum \tau\ =\ I\alpha\\\Rightarrow TR\ =\ (1)/(2)MR^2* (a)/(R)\\\Rightarrow a\ =\ (2T)/(M)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)`

From the eqn (1) and (2), we get,

`(mg\ -\ T)/(m)\ =\ (2T)/(M)\\\Rightarrow Mmg\ -\ MT\ =\ 2mT\\\Rightarrow Mmg\ =\ 2mT\ +\ MT\\\Rightarrow T\ =\ (Mmg)/(2m\ +\ M)\\\Rightarrow T\ = \ (8.0* 6.0* 9.81)/(2* 6.0\ +\ 8.0)\\\Rightarrow T\ =\ 23.54\ N`

Hence, the tension in the cord while the mass is falling is 23.54 N