A sphere mass m1 and a block of mass m2 are connected by a light cord that passes over a pulley. The radius of the pulley is R, and the mass of the thin rim is M.. The block slides on a frictionless, horizontal surface.
                Derive an expression for the linear acceleration of the two objects

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The expression for the linear expression of the two objects will be

[tex]a=\dfrac{m_1g}{m_2+m_1-M}[/tex]

What is acceleration?

Acceleration is defined as the change of the velocity with the time. Acceleration is a vector quantity and is defined by both the magnitude and the direction.

The mass of sphere is m1 radius is R and mass m2 is connected by light chord.

Now by using free body diagram

The tension in the chord will be

[tex]m_1g-T_1=m_1a[/tex]

[tex]T_1=m_1(g-a)[/tex]

For tension T2 we will have

[tex]T_2=m_2a[/tex]

Now the torque will be

[tex]T_{net}=I\alpha[/tex]

[tex]T_2R-T_1R=I(\dfrac{a}{R})[/tex]

[tex]T_2R-T_1R=MR^2(\dfrac{a}{R})[/tex]

[tex]T_2-T_1=Ma[/tex]

[tex]m_2a-m_1(g-a)=Ma[/tex]

[tex]m_2a-m_1g+m_1a=Ma[/tex]

[tex]a(m_2+m_1-M)=m_1g[/tex]

[tex]a=\dfrac{m_1g}{m_2+m_1-M}[/tex]

Thus the expression for the linear expression of the two objects will be

[tex]a=\dfrac{m_1g}{m_2+m_1-M}[/tex]

To know more about acceleration follow

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