Question

Each of the double pulleys shown has a centroidal mass moment of inertia of 0.25 kg-m2. an inner radius of 100 mm, and an outer radius of 150 mm. The mass m - 12.1 kg and the bearing friction is neglected Required information Determine the tension in the cord connecting the pulleys. (Round the final answer to two decimal places.) The tension in the cord connecting the pulleys is N.

Answer 1

For mass C;

ΙΣF= ma.

mg Тc = ma

TC mg - ma

For pulley A;

ΣΜΑ = ΙΑΠΑ

Ter – TR ΙΑ

=T=Tc-IA(R)

→T = (mg - ma) --IA GR) R

=T= mg ) - ma +) - IAR)

For pulley B;

MB = Igab

QAR Tr IB r

Tr IB aR r2

T = IB aR 73

= my (5) - ти (5) - (m) = le () qul

$\\\Rightarrow (12.1)(9.81)\left ( \frac{0.1}{0.15} \right ) - (12.1a)\left ( \frac{0.1}{0.15} \right ) - 0.25\left ( \frac{1}{0.1\times0.15} \right ) = 0.25\left ( \frac{a\times0.15}{0.1^3} \right )$

→ a= 1.27157 m/s

Tension in rope C;

$\Rightarrow T_C = 103.315\;\;N$

For pulley-A;

$T_C r - TR =I_A \left (\frac{a}{r} \right )$

$\Rightarrow 103.315 (0.1) - T(0.15) =(0.25) \left (\frac{1.27157}{0.1} \right )$

$\Rightarrow T = 47.684\;\;N$

$\Rightarrow T = 47.68\;\;N$

...(Answer)