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`        How can I solve this question: A parabolic bowl with its bottom at origin has the shape y =(x^2) /20. What is the maximum height where a bowl of mass m can be placed without slipping (x and y in metres, μ=0.5)?`
2 years ago

Vikas TU
8800 Points
```							Friction force would act tangentially obvious.Hence, f = umgNow, tanthetha = 2x/20 => x/10sinthetha = x/root(x^2 + 100)Hence,mgsin(thetha) = f = umg0.5 = x/root(x^2 + 100)x^2/(x^2 + 100) = ¼4x^2 = x^2 + 1003x^2 = 100x = 10/root(3) meter.
```
2 years ago
Amy
13 Points
```							dy/dx=2x/10 i.e. x/5 which is equal to tan thetaAlso umg cos theta = mg sin thetaI.e. u = tan theta Hence 1/2= X/5Therefore value of X=2.5 mPutting value of X in equation y=x^2/10We get. Y= 2.5*2.5/10Y= 0.625 m or 62.5 cm
```
2 months ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions