Velocity of Bead Relative to Wedge on a Smooth Hemispherical Surface

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SUMMARY

The discussion centers on calculating the velocity of a bead sliding on a smooth hemispherical wedge while the wedge itself moves. The key equations involved are the momentum conservation equation, \(MV=m(v\cos(\theta)-V)\), and the energy conservation equation, \(mgR(1-\cos(\theta))=(1/2)mv^2+(1/2)MV^2\). The initial attempt at solving these equations yielded an incorrect expression for velocity, \(v=\sqrt{\frac{2gR(1-\cos(\theta))(m+M)^2}{(M+m)^2+Mm(\cos(\theta))^2}}\). The error was identified as a misdefinition of \(v\), which should consistently represent the bead's velocity in terms of its components along the wedge.

PREREQUISITES
  • Understanding of classical mechanics principles, specifically conservation of momentum and energy.
  • Familiarity with the geometry of motion on curved surfaces.
  • Knowledge of trigonometric functions and their application in physics problems.
  • Ability to manipulate and solve algebraic equations involving multiple variables.
NEXT STEPS
  • Review the derivation of conservation of momentum in non-inertial reference frames.
  • Study the application of energy conservation in systems involving moving bodies on inclined planes.
  • Learn about the dynamics of motion on curved surfaces, focusing on hemispherical geometries.
  • Explore the use of vector components in defining motion in two dimensions.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding dynamics involving moving bodies on curved surfaces, particularly in the context of classical mechanics problems.

Sanchayan Dutta
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Homework Statement


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A bead of mass m kept at the top of a smooth hemispherical wedge of mass M and radius R is gently pushed towards right.As a result,the wedge slides due left.Find the magnitude of velocity of bead relative to the wedge.

Homework Equations


$$MV=m(v\cos(\theta)-V)$$
and,

$$mgR(1-cos(\theta))=(1/2)mv^2+(1/2)MV^2$$ i.e.
$$mgR(1-cos(\theta))=(1/2)mv^2+(1/2)M(\frac{mvcos(\theta)}{m+M})^2$$

The Attempt at a Solution



On solving the two equations I get $$v=\sqrt{\frac{2gR(1-cos(\theta))(m+M)^2}{(M+m)^2+Mm(cos(\theta))^2}}$$

But this answer is wrong according to my book.Where am I going wrong?
 
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How do you define v? You seem to use two different definitions in the first two equations.
 
mfb said:
How do you define v? You seem to use two different definitions in the first two equations.

Oh that's my mistake.So in the second equation i should use $$\sqrt{(vcos(\theta)-V)^2+(vsin(\theta))^2}$$ instead of v.Right?
 
That would make it consistent, yes.
 

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