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

Answer:

T = 23.54 N

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 = $\dfrac{1}{2}MR^2$

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

Hence, the angular acceleration of the pulley = $\alpha\ =\ \dfrac{a}{R}$

From the f.b.d. of the mass

$mg\ -\ T\ =\ ma\\\Rightarrow a\ =\ \dfrac{mg\ -\ T}{m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1),$

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

$\sum \tau\ =\ I\alpha\\\Rightarrow TR\ =\ \dfrac{1}{2}MR^2\times \dfrac{a}{R}\\\Rightarrow a\ =\ \dfrac{2T}{M}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)$

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

$\dfrac{mg\ -\ T}{m}\ =\ \dfrac{2T}{M}\\\Rightarrow Mmg\ -\ MT\ =\ 2mT\\\Rightarrow Mmg\ =\ 2mT\ +\ MT\\\Rightarrow T\ =\ \dfrac{Mmg}{2m\ +\ M}\\\Rightarrow T\ = \ \dfrac{8.0\times 6.0\times 9.81}{2\times 6.0\ +\ 8.0}\\\Rightarrow T\ =\ 23.54\ N$

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