Question

Ilustration 34.A fixed pulley of radius 20 cm and moment of inertia 0.32 kg.m2 about its axle has amassless cord wrapped around its rim. A 2 kg mass M is attached to the end ofthe cord. The pulley can rotate about its axis without any friction. Find theacceleration of the mass M. (Assume g = 10 m/s2) ​

Answer 1

m=2kg(mass of block)

r=0.2m

I=0.32kg

m2 Net force = mg-T=ma T=m(g-a)

T=20-2a

Torque=I@(alpha)

torque=T x r T x r= I @ (20-2a)0.2=0.32@ @=(20-2a)/1.6 r@=a

On putting value and solving we get a=2m/s2

Answer 2

The acceleration of the mass M is 2 m/s².

Explanation:

Given that,

Radius =20 cm

Moment of inertia = 0.32 kg m²

Mass = 2 kg

We need to calculate the net force

Using balance equation

$mg-T=ma$

$T=m(g-a)$...(I)

We need to calculate the acceleration

Using formula of torque

$\tau=F\times r$

We know that,

The torque is

$\tau=I\times \alpha$

$I\times\alpha=m(g-a)\times r$

$I\times\dfrac{a}{r}=m(g-a)\times r$

Put the value into the formula

$0.32\times\dfrac{a}{20\times10^{-2}}=2\times(10-a)\times20\times10^{-2}$

$1.6a=400\times10^{-2}-40a\times10^{-2}$

$(1.6+40\times10^{-2})a=400\times10^{-2}$

$a=\dfrac{400\times10^{-2}}{(1.6+40\times10^{-2})}$

$a=2\ m/s^2$

Hence, The acceleration of the mass M is 2 m/s².

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Topic : acceleration

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