a block with a mass of m=10kg is connected to a pulley with a mass of M=2.5kg and a radius of R=0.5m. what is(adsbygoogle = window.adsbygoogle || []).push({});

1) the angular acceleration of the pulley

2) the acceleration of the block

3) the tension in the rope connecting them?

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equations

F=ma

I[tex]\alpha[/tex]=|FR|sin

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

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my attempt

1)I[tex]\alpha[/tex]=|FR|sin

the angle between the radius and the force (the rope) is 90 degrees, sin90=1

I[tex]\alpha[/tex]=|TR|

I=0.5MR[tex]^{2}[/tex]

using newtons 2nd law on the block mg-T=ma,=> T=m(g-a)

[tex]\alpha[/tex]=|TR|/I

=[tex]\frac{mR(g-a)}{0.5MR^{2}}[/tex]

=[tex]\frac{2m(g-a)}{MR}[/tex]

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

[tex]\alpha[/tex]=[tex]\frac{2m(g-[tex]\alpha[/tex]R)}{MR}[/tex]

[tex]\alpha[/tex]MR=2mg-2m[tex]\alpha[/tex]R

[tex]\alpha[/tex](MR+2mR)=2mg

[tex]\alpha[/tex]=[tex]\frac{2mg}{R(M+2m)}[/tex]

plug in the numbers, and i get [tex]\alpha[/tex]= [tex]\frac{160}{9}[/tex] rad/s[tex]^{2}[/tex]

2) to find the acceleration of the block, a

a=[tex]\alpha[/tex]R=[tex]\frac{2mg}{(M+2m)}[/tex]=[tex]\frac{80}{9}[/tex]m/s[tex]^{2}[/tex]

3) to find the tension in the rope, using newtons 2nd law on the block mg-T=ma,=> T=m(g-a) using the a i found in 2)

T=m(g-[tex]\frac{2mg}{(M+2m)}[/tex])=100/9 N

are these workings correct??

the answers are all correct except the answer for 3) which my book says T=800/9

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# Homework Help: Angular acceleration of the pulley

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