Question #105197
7. A mass (M1 = 5 kg) is connected by a light cord to a mass (M2 = 4 kg) which slides on a smooth surface, as shown in the figure. The pulley (radius=0.20 m) rotates about a frictionless axle. The acceleration of M2 is 3.5 m/s2. What is the moment of inertia of the pulley?

8. A playground merry-go-round has a disk-shaped platform that rotates with negligible friction about a vertical axis. The disk has a mass of 200 Kg and a radius of 2 m. A 40-Kg child rides at the center of the merry-go-round while a playmate sets it turning at 0.35 rev/sec. If the child then walks along a radius to the outer edge of the disk, how fast will the disk be turning? (HINT: Treat the child as a particle)
1
Expert's answer
2020-03-13T11:05:22-0400

As per the given question,

mass of the object (M1)=5kg(M_1)=5 kg

mass of the second object (M2)=4kg(M_2)=4kg

Radius of the pulley (R)=0.2m

Acceleration of the object (a)=3.5m/sec2(a)=3.5 m/sec^2




Now,

Let the tension in the string between the pulley and block M1M_1 is T1T_1 and tension in the string between the pulley and block M2M_2 is T2T_2

So, M1gT1=M1a(i)M_1g-T_1=M_1a------(i)

T2=M2a(ii)T_2= M_2a--------(ii)

Now from the torque equation on the pulley

(T1T2)R=Iα(iii)(T_1-T_2)R=I\alpha-----(iii)

Now, substituting the given value in the equation (ii)

T2=4×3.5=14.0NT_2=4\times 3.5= 14.0N

Now, substituting the given values in the equation (i)

5×9.8T2=5×3.55\times 9.8-T_2=5\times 3.5

T1=31.5N\Rightarrow T_1=31.5N

Now, substituting the values in the equation (iii)

(31.514)×0.2=IaR\Rightarrow (31.5-14)\times 0.2=I\dfrac{a}{R}

I=17.5×0.2×0.23.5=0.2kgm2\Rightarrow I =\dfrac{17.5\times 0.2\times0.2}{3.5}=0.2 kg-m^2


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