A thin uniform rod of length and certain mass is kept on a frictionless horizontal table with a massless string of length fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O. [2024]

Question

A thin uniform rod of length $L$ and certain mass is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O. If a horizontal impulse P is imparted to the rod at a distance $x = \frac{L}{n}$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point O, with the rod remaining aligned with the string. In such a case, the value of $n$ is _______. [2024]

Smart Achievers · Accepted Answer

(18)

$Linear impulse, \int F d t = Δ momentum = m (V_{c m} - 0)$

$or, P = m (ω r_{c m}) = m ω (L + \frac{L}{2})$

$or, P = m ω (\frac{3 L}{2}) \dots (i)$

$And angular impulse \int τ d t = Δ angular momentum$

$\int r \times F d t = Δ L \Rightarrow r \times \int F d t = I (ω - 0) Here I = moment of inertia about axis of rotation.$

$(L + \frac{L}{2} + x) P = (I_{c m} + m d^{2}) ω$

$= (\frac{m L^{2}}{12} + m {(L + \frac{L}{2})}^{2}) ω$

$(\frac{3 L}{2} + x) P = m L^{2} (\frac{1}{12} + {(\frac{3}{2})}^{2}) ω$

$(\frac{3 L}{2} + x) P = m L^{2} (\frac{7}{3}) ω \dots (i i)$

$Dividing equation (ii) by (i)$

$(\frac{3 L}{2} + x) = \frac{L (\frac{7}{3})}{(\frac{3}{2})} \Rightarrow \frac{3 L}{2} + x = L (\frac{14}{9})$

$or, x = \frac{L}{18}$ $∴$ $n = 18$

Solution

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