Question

A yo-yo with a mass of 0.075 kg and a rolling radius of 2.50 cm (the distance from the axis of the pulley to where the string comes off the spool) rolls down a string with a linear acceleration of 6.50 m/s2. Approximate the rotational inertia of the yo-yo with that of disk with mass, m, and radius, r, rotating about its center (mr2/2). Calculate the tension in the string.

Answer 1

Answer:

0.24825 N

0.0000238701923077 kgm²

Explanation:

m = Mass of yo yo = 0.075 kg

a = Acceleration = 6.5 m/s²

g = Acceleration due to gravity = 9.81 m/s²

The net force is given by

$F_n=mg-T$

$\Rightarrow T=mg-ma$

$\Rightarrow T=m(g-a)$

$\Rightarrow T=0.075(9.81-6.5)$

$\Rightarrow T=0.24825\ N$

The tension in the string is 0.24825 N

Angular acceleration is given by

$\alpha=\dfrac{a}{r}\\\Rightarrow \alpha=\dfrac{6.5}{2.5\times 10^{-2}}\\\Rightarrow \alpha=260\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow Tr=I\alpha\\\Rightarrow I=\dfrac{Tr}{\alpha}\\\Rightarrow I=\dfrac{0.24825\times 2.5\times 10^{-2}}{260}\\\Rightarrow I=0.0000238701923077\ kgm^2$

The moment of inertia is 0.0000238701923077 kgm²