A uniform disc of radius R and mass M is mounted on an axis supported in fixed frictionless bearing. A light chord is wrapped around the rim of the wheel and suppose that we hang a body of mass m from the chord. Find the angular acceleration of the disc and tangential acceleration of point on the rim.
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The situation is shown in the fig.
Let f be the tension in the chord. Now, $\text{mg}-\text{T}=\text{ma},\dots(1)$ where a is the tangential acceleradon of a point
on the rim of the disc We know that $\tau=\text{l}\alpha.$ But the resultant torque on the disc = TR and
the rotational inertia $\text{l}=\frac{1}{2}\text{MR}^2$ $\therefore\text{TR}=\frac{1}{2}\text{MR}^2\Big(\frac{\text{a}}{\text{R}}\Big)(\because\alpha=\frac{\text{a}}{\text{R}})$ or $2\text{TR}=\text{Ma}$ or $\text{a}=\frac{2\text{T}}{\text{M}}\dots(2)$ From equations (r) and (ir), we get $\text{mg}-\Big(\frac{\text{Ma}}{2}\Big)=\text{ma}$ or $\text{a}=\Big(\frac{2\text{m}}{\text{M+2M}}\Big)\text{g}\dots(3)$ Again,$\text{mg}-\text{T}=\text{m}\times\Big(\frac{2\text{T}}{\text{M}}\Big)$ or $\text{T}=\Big(\frac{\text{mM}}{\text{M}+2\text{m}}\Big)\text{g}\dots(4)$
Let f be the tension in the chord. Now, $\text{mg}-\text{T}=\text{ma},\dots(1)$ where a is the tangential acceleradon of a point
on the rim of the disc We know that $\tau=\text{l}\alpha.$ But the resultant torque on the disc = TR and
the rotational inertia $\text{l}=\frac{1}{2}\text{MR}^2$ $\therefore\text{TR}=\frac{1}{2}\text{MR}^2\Big(\frac{\text{a}}{\text{R}}\Big)(\because\alpha=\frac{\text{a}}{\text{R}})$ or $2\text{TR}=\text{Ma}$ or $\text{a}=\frac{2\text{T}}{\text{M}}\dots(2)$ From equations (r) and (ir), we get $\text{mg}-\Big(\frac{\text{Ma}}{2}\Big)=\text{ma}$ or $\text{a}=\Big(\frac{2\text{m}}{\text{M+2M}}\Big)\text{g}\dots(3)$ Again,$\text{mg}-\text{T}=\text{m}\times\Big(\frac{2\text{T}}{\text{M}}\Big)$ or $\text{T}=\Big(\frac{\text{mM}}{\text{M}+2\text{m}}\Big)\text{g}\dots(4)$
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