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By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and m2 = 43.7 kg. The pulley can be treated as a uniform, solid, cylindrical disk. The downward acceleration of the 43.7 kg block is observed to be exactly one-half the acceleration due to gravity. Noting that the tension in the rope is not the same on each side of the pulley, find the mass of the pulley.

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lublana

Answer:

14.8 kg

Explanation:

We are given that

[tex]m_1=43.7 kg[/tex]

[tex]m_2=12.1 kg[/tex]

[tex]g=9.8 m/s^2[/tex]

[tex]a=\frac{1}{2}(9.8)=4.9 m/s^2[/tex]

We have to find the mass of the pulley.

According to question

[tex]T_2-m_2 g=m_2 a[/tex]

[tex]T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N[/tex]

[tex]T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N[/tex]

Moment of inertia of pulley=[tex]I=\frac{1}{2}Mr^2[/tex]

[tex](T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)[/tex]

Where [tex]\alpha=\frac{a}{r}[/tex]

[tex](177.87-214.13)=-\frac{1}{2}(4.9)M[/tex]

[tex]-36.26=-\frac{1}{2}(4.9)M[/tex]

[tex]M=\frac{36.26\times 2}{4.9}=14.8 kg[/tex]

Hence, the mass of the pulley=14.8 kg

The mass of the pulley is 14.8 kg.

Calculation of the mass:

Since

m1 = 12.1 kg

m2 = 43.7 kg

g = 9.8 m/s^2

a = 1/2(9.8) = 4.9m/s^2

Now we know that

T2 - m2g = m2a

T2 = m2a + m2g

= m2(a + g)

= 12.1(9.8 + 4.9)

= 177.87 N

Now

T1 = m1(g - a)

= 43.7 (9.8-4.9)

= 214.13 N

Now moment of inertia should be

(T2 - T1)r = -1/2(4.9M)

(177.87-214.13) = -1.2(4.9)M

-36.26 = -1.2(4.9)M

M = 14.8 kg

hence, The mass of the pulley is 14.8 kg.

Learn more about mass here: https://brainly.com/question/18470727.

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