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A mass of 5.00 kg pulls down vertically on a string that is wound around a rod of radius 0.100 m and negligible moment of inertia. The rod is fixed in the center of a disk. The disk has mass 125 kg and radius 0.2 m. They turn freely about a fixed axis through the center. What is the angular acceleration of the rod, in radians/s?

a. 2
b. 25
c. 50
d. 75
e. 125

Respuesta :

Answer:

The angular acceleration of the rod is 2 rad/s².

(a) is correct option.

Explanation:

Given that,

Mass of string = 5.00 kg

Radius = 0.100 m

Mass of disk = 125 kg

Radius of disk = 0.2 m

We need to calculate the acceleration

Using balance equation

[tex]mg-T=ma[/tex]

Put the value of m

[tex]5g-T=5a[/tex]....(I)

We need to calculate the tension

Using balance equation

[tex]T\times r=I\times \alpha[/tex]

[tex]T=\dfrac{I\times\alpha}{r}[/tex]

[tex]T=\dfrac{\dfrac{mr^2}{2}\times\alpha}{r}[/tex]

[tex]T=\dfrac{\dfrac{mr^2}{2}\times\dfrac{v}{r}}{r}[/tex]

Put the value into the formula

[tex]T=\dfrac{\dfrac{125\times(0.2)^2}{2}\times a}{(0.1)^2}[/tex]

[tex]T=250a[/tex]....(II)

From equation (I) and (II)

[tex]255a=5g[/tex]

Put the value into the formula

[tex]a=\dfrac{5\times9.8}{255}[/tex]

[tex]a=0.2\ m/s^2[/tex]

We need to calculate the  angular acceleration of the rod

Using formula of angular acceleration

[tex]\alpha=\dfrac{a}{R}[/tex]

Put the value into the formula

[tex]\alpha=\dfrac{0.2}{0.1}[/tex]

[tex]\alpha=2\ rad/s^2[/tex]

Hence, The angular acceleration of the rod is 2 rad/s².

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