A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.

What is the period if the rod and clay swing as a pendulum?

Respuesta :

Answer:

0.76 s

Explanation:

We are given that

Length of rod,L=20 cm=[tex]\frac{20}{100}=0.20m[/tex]

1 m=100 cm

Mass of rod,M=190 g=[tex]\frac{190}{1000}=0.19kg[/tex]

Mass of ball,m=19 g=[tex]\frac{19}{1000}=0.019 kg[/tex]

Using 1 kg=1000g

We have to find the period if the rod and clay swing as a  pendulum.

Moment of inertia of rod-clay=Moment of inertia of rod+moment of inertia of clay

[tex]I_{rod-clay}=I_{rod}+I_{clay}[/tex]

[tex]I_{rod-clay}=\frac{1}{3}ML^2+mL^2[/tex]

Substitute the values then we get

[tex]I_[rod-clay}=\frac{1}{3}(0.19)(0.20)^2+(0.019)(0.20)^2[/tex]

[tex]I_{rod-clay}=3.29\times 10^{-3} kgm^2[/tex]

Now, the center of mass of the combination of the rod and clay is given by

[tex]y=\frac{Md_1+md_2}{M+m}[/tex]

Substitute [tex]d_1=\frac{L}{2}[/tex]=Distance between pivot and the center of the rod

[tex]d_2=L=[/tex]The distance  between rod and clay

Using the formula

[tex]y=\frac{0.19\times \frac{0.20}{2}+0.019\times 0.20}{0.19+0.019}[/tex]

[tex]y=0.1091[/tex] m

Time period of the oscillation of the system of the rod and the clay is given by

[tex]T=2\pi\sqrt{\frac{I_{rod-clay}}{(M+m)yg}}[/tex]

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

Using the formula

Time period=[tex]2\times 3.14\sqrt{\frac{3.29\times 10^{-3}}{(0.19+0.019)\times 9.8\times 0.1091}}[/tex]

Time-period=0.76 s

Hence, the period =0.76 s

This question involves the concepts of the time period, the moment of inertia, and the center of mass.

The time period of the rod and the clay swing as a pendulum is "0.759 s".

The time period of the rod and the clay swing as a pendulum is given by the following formula:

[tex]T=2\pi \sqrt{\frac{I}{(m_1+m_2)yg}}[/tex]

where,

T = time period = ?

m₁ + m₂ = mass of the system = mass of rod + mass of clay

m₁ + m₂ = 190 g + 19 g = 209 g = 0.209 kg

g = acceleration due to gravity = 9.81 m/s²

y = center of mass of the rod and clay system

y = [tex]\frac{m_1d_1+m_2d_2}{m_1+m_2}=\frac{(0.19\ kg)(0.1\ m)+(0.019\ kg)(0.2\ m)}{0.209\ kg} = 0.109\ m[/tex]

I = moment of inertia of the rod and clay system

I = moment of inertia of rod + moment of inertia of clay

I = [tex]\frac{1}{3}m_1L^2+m_2L^2 = \frac{1}{3}(0.19\ kg)(0.2\ m)^2+(0.019\ kg)(0.2\ m)^2[/tex]

I = 0.0025 kg.m² + 0.00076 kg.m² = 0.00326 kg.m²

Therefore,

[tex]T=2\pi \sqrt{\frac{0.00326\ kg.m^2}{(0.209\ kg)(0.109\ m)(9.81\ m/s^2)}}[/tex]

T = 0.759 s

Learn more about the time period of a pendulum here:

https://brainly.com/question/16797678?referrer=searchResults

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