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A uniform cylinder of radius `r` and mass `m` can rotate freely about a fixed horizontal axis. A thin cord of length l and mass `m_(0)` is would on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length `x` of the hanging part of the end. the wound part of the cord is supposed to have its centre of gravity on the cylinder axis is shown in figure. |
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Answer» Let us use the equation `(dM_z)/(dt)=N_z` relative to the axis through O (1) For this purpose, let us find the angular momentum of the system `M_z` about the given rotation axis and the corresponding torque `N_x`. The angular momentum is `M_z=Iomega+mvR=(m_0/2+m)R^2omega` [where `I=(m_0)/(2)R^2` and `v=omegaR` (no cord slipping)] So, `(dM_z)/(dt)=((MR^2)/(2)+mR^2)beta_z` (2) The downward pull of gravity on the overhanging part is the only external force, which exertes a torque about the z-axis, passing through O and is given by, `N_z=(m/l)xgR` Hence from the equation `(dM_z)/(dt)=N_z` `((MR^2)/(2)+mR^2)beta_z=m/lxgR` Thus, `beta_z=(2mgx)/(lR(M+2m))gt0` Note: We may solve this problem using conservation of mechanical energy of the system (cylinder + thread) in the uniform field of gravity. |
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