Question

The two blocks in FIGURE CP12.86 are connected by a massless rope that passes over a pulley. The pulley is 12 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.50 N m. If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Short answer

Time taken to reach the floor is 0.857 sec

Step-by-step solution

Given information

Mass of block m₂=2 kg
Acceleration due to gravity =9.8 m/s²
Mass of the pulley =1.2 Kg
Radius of the pulley (r)=5.0 cm=0.05 m

Explanation

First write equation of motion by force on the system

$\sum F=mg-T=ma$

Force exerted by the cord

$$\begin{gathered} \Rightarrow \sum F=mg=ma+T \\ \Rightarrow \left(m_1+m_2+\frac{I}{R^2}\right)a=\left(m_1-m_2\right)g.................................\left(1\right) \end{gathered}$$

Inertia of pulley is

$$\begin{gathered} I=\frac{M{r}^2}{2} \\ \frac{I}{r^2}=\frac{M}{2}....................................\left(2\right) \end{gathered}$$

From equation(1) and (2), we get acceleration as

$$\begin{gathered} a=\frac{\left(m_1-m_2\right)}{\left(m_1+m_2+\frac{M}{2}\right)}g \\ a=\frac{(4\mathrm{kg}-2\mathrm{kg})}{(4\mathrm{kg}+2\mathrm{kg}+1.2/2\mathrm{kg})}9.8\frac{\mathrm{m}}{{\mathrm{s}}^2} \\ \Rightarrow a=2.72\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

From the equation of motion we get

$$\begin{gathered} h=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow t=\sqrt{\frac{2h}{a}} \end{gathered}$$

substitute values

$\Rightarrow t=\sqrt{\frac{2\left(1m\right)}{2.72m{s}^{-2}}}=0.857\mathrm{s}$