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In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2, and that counterclockwise is considered the positive rotational direction.a) Assume that the mass of the swing bar, as shown in the figure, is negligible. Find the magnitude of the angular acceleration a(alpha) of the seesaw. Express in variables of m1, m2, l and g.b) Now consider a similar situation, except that now the swing bar itself has mass mbar . (Part C figure)
Find the magnitude of the angular acceleration a (alpha) of the seesaw. GIve answer as variables as before.

Respuesta :

Answer:

Part a)

[tex]\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}[/tex]

Part b)

[tex]\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}[/tex]

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

[tex]\tau = I\alpha[/tex]

here we will have

[tex]\tau = (m_1g - m_2g)(\frac{L}{2})[/tex]

now we will have inertia of two masses given as

[tex]I = (m_1 + m_2)(\frac{L}{2})^2[/tex]

now we have

[tex]I = (m_1 + m_2)\frac{L^2}{4}[/tex]

now the angular acceleration is given as

[tex]\alpha = \frac{\tau}{I}[/tex]

so we have

[tex]\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}[/tex]

Part b)

Now if the rod is not massles then we will have total inertia given as

[tex]I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}[/tex]

so we will have

[tex]I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}[/tex]

now the acceleration is given as

[tex]\alpha = \frac{\tau}{I}[/tex]

[tex]\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}[/tex]

snehashish65

(A) The magnitude of angular acceleration of the seesaw in this case is [tex]\dfrac{ 2(m_{1}-m_{2})g}{ (m_{1}+m_{2})L}[/tex].

(B) The magnitude of angular acceleration of the seesaw in this case is

[tex]\dfrac{ (m_{1}g-m_{2}g) }{L((m_{1}+m_{2})+\dfrac{1}{3})}[/tex].

(A)

For a massless sea saw bar, with attached masses at each end, the torque produced due to the masses is,

[tex]\tau = (m_{1}g-m_{2}g)\dfrac{L}{2}[/tex] .................................................(1)

And, moment of inertia of the system of two masses is,

[tex]I= (m_{1}+m_{2})\dfrac{L^{2}}{4}[/tex]     ................................................(2)

The using the expression of torque as,

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

Substituting the values as,

[tex]\alpha=\dfrac{ (m_{1}g-m_{2}g)\dfrac{L}{2}}{ (m_{1}+m_{2})\dfrac{L^{2}}{4}}\\\alpha=\dfrac{ 2(m_{1}-m_{2})g}{ (m_{1}+m_{2})L}[/tex]

Thus, the magnitude of the angular acceleration of the seesaw is [tex]\dfrac{ 2(m_{1}-m_{2})g}{ (m_{1}+m_{2})L}[/tex].

(B)

For massless rod, the total moment of inertia is given as,

[tex]I_{total}=I+\dfrac{mL^{2}}{12} \\I_{total}=(m_{1}+m_{2})\dfrac{L^{2}}{4}+\dfrac{mL^{2}}{12}[/tex]

Here, m is the mass of sea-saw bar.

[tex]I_{total}=\dfrac{L^2}{4} ((m_{1}+m_{2})+\dfrac{1}{3})[/tex]

Then the angular acceleration is,

[tex]\tau = I_{total} \times \alpha\\ \alpha=\dfrac{\tau }{I_{total}}\\ \alpha=\dfrac{ (m_{1}g-m_{2}g)\dfrac{L}{2} }{\dfrac{L^2}{4} ((m_{1}+m_{2})+\dfrac{1}{3})}\\\alpha=\dfrac{ (m_{1}g-m_{2}g) }{L((m_{1}+m_{2})+\dfrac{1}{3})}[/tex]

Thus, the angular acceleration of the sea-saw is [tex]\dfrac{ (m_{1}g-m_{2}g) }{L((m_{1}+m_{2})+\dfrac{1}{3})}[/tex].

Learn more about moment of inertia and torque here:

https://brainly.com/question/22687691?referrer=searchResults

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