(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

http://session.masteringphysics.com/problemAsset/1070551/3/12.P70.jpg

The two blocks in the figure are connected by a massless rope that passes over a pulley. The pulley is 15 cm in diameter and has a mass of 2.6 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.46 N-m.

If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

2. Relevant equations

T_{net}=I[tex]\alpha[/tex]

T=FD

F=ma

[tex]\alpha[/tex]=a/r

Moment of inertia of a disc:

I=.5mr^{2}

g is gravity!

3. The attempt at a solution

T1 is tension of the left mass, T2 is tension of the right mass

m1 is mass of left mass, m2 is mass of right mass

T1=-m1a+m1g

T2=m2a+m2g

[tex]\Sigma[/tex]T= T_{T1}- T_{Pully friction}- T_{T2}

I[tex]\alpha[/tex]=T_{T1}- T_{Pully friction}- T_{T2}

Ia/r=(-m1a+m1g)r - .46 - (m2a+m2g)r

Ia/r= -m1ar + m1gr - .46 -m2ar -m2gr

Ia = -m1ar^{2}+ m1gr^{2}- .46r -m2ar^{2}-m2gr^{2}

Ia + m1ar^{2}+ m2ar^{2}= m1gr^{2}- .46r -m2gr^{2}

a(I + m1r^{2}+ m2r^{2}) = m1gr^{2}- .46r -m2gr^{2}

a = (m1gr^{2}- .46r -m2gr^{2}) / (I + m1r^{2}+ m2r^{2})

I=.5mr^{2}

I = (.5)(2.6)(.075)^{2}

I = .0073125

m1 = 4kg

m2 = 2kg

r = .075m

g=9.8

a = ((4)(9.8)(.075^{2}) - .46(.075) -(2)(9.8)(.075)^{2}) / (.0073125) + (4)(.075)^{2}+ (2)(.075)^{2}

a = (.2205 - .0345 - .11025) / (.0073125 + .0225 + .01125)

a = .07575/.0410625

a = 1.8447

yf=yi+vit+.5at^2

0=1 + 0 + .92235t^2

t=1.04124 seconds

is this correct?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: 2 hanging masses on Pully with friction time to reach floor?

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