One end of massless spring of spring constant $ 100 \,N/m $ and natural length $ 0.49\, m $ is fixed and other end is connected to a body of mass $ 0.5 \,kg $ lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of $ 2 $ rad/s, then the elongation of the spring will be
Given, $M=0.5\,kg$, $\omega=2\, rad/ s$, $l=0.49\, m$, and $k=100 \,N/ m $ Figure represents situation, as given in question
Centripetal force on the blocks = Spring force $\Rightarrow Mr\omega^{2}=k \Delta x $ $\Rightarrow 0.5\left(l+\Delta x\right)\left(2\right)^{2}=100\cdot\Delta x$ $\Rightarrow \left(0.49+\Delta x\right)4=100\cdot\Delta x\times2$ $\Rightarrow 0.49+\Delta x=\frac{200}{4} \Delta x $ $\Rightarrow 0.49+\Delta x=50\Delta x$ $\therefore 49 \Delta x=0.49$ $\Delta x=0.01\, m = 1\,cm$
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Top Questions on System of Particles & Rotational Motion
The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
The distance between the pair of particles in such a body does not replace or alter. Rotational motion can be described as the motion of a rigid body originates in such a manner that all of its particles move in a circle about an axis with a common angular velocity.
The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.
