Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionDynamicsHard2 minQB

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Passage / Comprehension

A very small cube of mass $2 k g$ is placed on the surface of a funnel as shown in the figure. The funnel is rotating about its vertical axis of symmetry with angular velocity $ω$ . The wall of the funnel makes an angle $3 7^{\circ}$ with the horizontal. The distance of the cube from the axis of rotation is $20 c m$ , and the frictional coefficient is $μ$ . (Take $g = 10 m / s^{2}$ )

The minimum value of angular velocity for which no relative slipping occurs is : $(μ = 2/3)$ :-

Options:

Answer:

Solution:

For minimum value of ' $ω$ ', cube has tendency to slip downwards, so in order to prevent slipping, friction will act up the plane.

$(f)_{max} = μ N = [$ minimum $ω$ is to be found $]$

From FBD of cube :

$N cos 3 7^{\circ} + μ N cos 5 3^{\circ} = mg$ ..........(1)

$N sin 3 7^{\circ} - μ N cos 3 7^{\circ} = m ω^{2} r$ ............(2)

From (1) : $\frac{4 N}{5} + \frac{2}{3} N \times \frac{3}{5} = 20$

$6 N = 20 \times 5 \Rightarrow N = \frac{100}{6} N$

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Dynamics

⏱ 2mℹ️ Source: QB

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