Question

In the given arrangement, pulley is free to rotate about its center and has moment of inertia / = 4m x R2, where Ris radius of pulley. The string is wrapped over the pulley and does not slip. The acceleration of block of mass m is

1, R

m

2m​

Answer 1

Answer:

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Answer 2

Answer:

The acceleration of the block of mass m will be $(8g/7)$ i.e.11.42$ms^{-2}$.

Explanation:

As per the question,

m1=2m

m2=m

Moment of inertia of the disc(I)=4m ×$R^{2}$

Tension in both cables will be T1 and T2

The equation of tension for both the masses,

2mg-T1=2ma

T1=2(mg-ma)                   (1)

T2-mg=ma

T2=(mg+ma)                   (2)

Torque on the pulley,

(T1-T2)R=Iα            (3)

I=moment of inertia

R=radius of the disc

α=angular acceleration

by using equations (1), (2), and (3) we get:

(2mg-2ma-mg-ma)R=Iα

(mg-3ma)R=Iα

Relation between acceleration and angular acceleration is;

α=a$/$R              (4)

a=linear acceleration

(mg-3ma)R=4m$R^{2}$×a/R

g-3a=4a

g=7a

a=g$/$7

g=acceleration due to gravity=10$ms^{-2}$

The acceleration of block of mass m will be,

a'=a+g

Now let's substitute the value of a and g in the above equation,

a'=g$/$7 +g=$8g/7$=80/7=11.42$ms^{-2}$

Thus, we conclude that the acceleration of block of mass m will be $8g/7$i.e.11.42$ms^{-2}$.