Answer to Question #105196 in Classical Mechanics for Samson

Question

5. A 15-kg object and a 10-kg object are suspended, joined by a cord that passes over a pulley with a radius of 15 cm and a mass of 3 kg. The cord has a negligible mass and does not slip on the pulley. Treat the pulley as a uniform disk, and determine the linear acceleration of the two objects after the objects are released from rest.

6. A block of mass m1 = 2 kg and a block of mass m2 = 6 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30 degrees. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks, and (b) the tensions in the string on both sides of the pulley.

Answer 1

As per the given question,

masses of the objects are $(m_1)=15 kg$

and $(m_2) =10kg$

mass of the pulley =3 kg

Radius of the pulley =15 cm= 0.15 m

now applying newton's law,

Let it is moving with a acceleration and 15kg block is hanged left side and 10 is hanged to right side of the pulley, Let the tension in the left side of the string is $T_1$ and right side of the string is $T_2$

So, $15g-T_1=15a-------(i)$

$T_2-10g=10a-------(ii)$

from equation (i) and (ii)

$15g-10g-T_1-T_2=15a-10a$

$5g-(T_1-T_2)=5a$

$T_1-T_2=5g-5a--------(iii)$

torque on the pulley,

$(T_1-T_2)R=I\alpha ----------(iv)$

Now, from the equation (iii) and (iv)

$\Rightarrow (5g-5a)R=\dfrac{mR^2}{2}\times\alpha$

$\Rightarrow 5g-5a=\dfrac{3a}{2}$

$\Rightarrow 5g= 5a+\dfrac{3a}{2}$

$\Rightarrow \dfrac{13a}{2}=5g$

$a=\dfrac{10g}{13} m/sec^2$