Question
The chain of mass $\rho$ per unit length passes over the small freely turning pulley and is released from rest with only a small imbalance $h$ to initiate motion. Determine the acceleration $a$ and velocity $v$ of the chain and the force $R$ supported by the hook at $A,$ all in terms of $h$ as it varies from essentially zero to $H$. Neglect the weight of the pulley and its supporting frame and the weight of the small amount of chain in contact with the pulley. (Hint: The force $R$ does not equal two times the equal tensions $T$ in the chain tangent to the pulley.)
Step 1
Let's denote the length of the chain hanging on the left side of the pulley as \( x \) and the length on the right side as \( H - x \), where \( H \) is the total length of the chain. The imbalance \( h \) is defined such that \( x = \frac{H}{2} + h \). The mass Show more…
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A solid, frictionless cylindrical pulley has a radius R = 0.550 m. A block is attached to a cord that is wrapped around the outside of the pulley. The block is falling with a downward acceleration of 3.80 m/s2, and the tension in the cord is 0.850 N. The free body diagrams for the pulley and the block are shown to the right. Note that the magnitudes of the two tensions T1 and T2 are equal. (Just call the tension T.) Also, "n" in the free body diagram of the pulley represents the normal force provided by the axle of the pulley, equal in magnitude to the weight of the pulley. Note that the normal force and the weight force do not produce torques on the pulley. (Why is this?)
First of all, it is clear that the chain does not move in the vertical direction during the uniform rotation. This means that the vertical component of the tension $T$ balances gravity. As for the horizontal component of the tension $T$, it is constant in magnitude and permanently directed toward the rotation axis. It follows from this that the C.M. of the chain, the point $C$, travels along horizontal circle of radius $\rho$ (say). Therefore we have, $$ T \cos \theta=m g \text { and } T \sin \theta=m \omega^{2} \rho $$ Thus $\quad \rho=\frac{g \tan \theta}{\omega^{2}}=0 \cdot 8 \mathrm{~cm}$ and $\quad T=\frac{m g}{\cos \theta}=5 \mathrm{~N}$
Physical Fundamentals Of Mdchanics
Laws of Conservation of Energy, Momemtum, and Angular Momentum
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