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A Block of Mass M is Kept on a Horizontal Ruler. the Friction Coefficient Between the Ruler and the Block is μ. the Ruler is Fixed at One End and the Block is at a Distance L from the Fixed End. - Physics

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प्रश्न

A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is μ. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip? 

योग

उत्तर

(a) Given:
Mass of the block = m
Friction coefficient between the ruler and the block = μ
Let the maximum angular speed be ω1 for which the block does not slip
Now, for the uniform circular motion in the horizontal plane, we have : 

\[\mu \text{mg = m }\omega_1^2 L\]

\[ \therefore \omega_1 = \sqrt{\frac{\mu g}{L}}\]

(b) Let the block slip at an angular speed ω2. 
For the uniformly accelerated circular motion, we have :

\[\mu \text{mg} = \sqrt{\left( \text{ m }\omega_2^2 L \right)^2 + \left( \text{mL } \alpha^2 \right)^2}\]

\[ \Rightarrow \omega_2^4 + \alpha^2 = \frac{\mu^2 g^2}{L^2}\]

\[\mu \text{mg } = \sqrt{\left( \text{ m }\omega_2^2 L \right)^2 + \left( \text{ mL } \alpha^2 \right)^2}\]

\[ \Rightarrow \omega_2^4 + \alpha^2 = \frac{\mu^2 g^2}{L^2}\]

\[\Rightarrow \omega_2 = \left[ \left( \frac{\mu g}{L} \right)^2 - \alpha^2 \right]^{1/4}\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Circular Motion - Exercise [पृष्ठ ११५]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 7 Circular Motion
Exercise | Q 21 | पृष्ठ ११५

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