Question

Two blocks are connected by a massless rope that passes over a 1 kg pulley with a radius of 12 cm. The rope moves over the pulley without slipping. The mass of block A is 2.1 kg and the mass of block B is 4.1 kg. Block A is also connected to a horizontally-mounted spring with a spring constant of 358 J/m2. What is the angular frequency (in rad/s) of oscillations of this system

Answer 1

Answer:

$F=1.159$

Explanation:

From the question we are told that:

Mass of pulley $M=1kg$

Radius $r=12cm$

Mass of block A $M_a=2.1kg$

Mass of block B $m_b=4.1kg$

Spring constant$\mu= 358 J/m2$

Generally the equation for Torque is mathematically given by

Since $\sumF=ma$

At mass A

 $T_2-f_3=2.1a$

At mass B

 $4.8-T_1=4.1a$

At  Pulley

 $R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}$

 $R(T_1-T_2)=0.55a$

Therefore the equation for total force F

At mass A+At mass B+At  Pulley

 $(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a$

 $(T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a$

 $-f_3+4.1=6.75a$

 $-f_3=6.75a+4.8$

Since From above equation

$M_{eff}=6.7kg$

Therefore

$T=2\pi \sqrt{{\frac{M_{eff}}{k}}$

$T=2\pi \sqrt{{\frac{6.75}{\mu}}$

$T=0.862s$

Generally the equation for frequency is mathematically given by

$F=\frac{1}{T} \\F=\frac{1}{0.862}$

$F=1.159$