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        A particle with mass m begins to slide down a fixed smooth sphere from top. What is its tangential acceleration when it breaks off the sphere
one year ago

Khimraj
3008 Points
							if a particle starts from the topthe speed of the particle will be given by energy conservation lawit is given aschange in the potential energy = gain in kinetic energy$mgR(1 - cos\theta) = \frac{1}{2} mv^2$from force equation nowsince it break off at this position so normal force will be zero$mgcos\theta = \frac{mv^2}{R}$now we will use the value from above two equations$mgcos\theta = 2mg*(1 - cos\theta)$by solving above equation$3cos\theta = 2$$cos\theta = \frac{2}{3}$ now for tangential acceleration as we know that $a_t = \frac{F_t}{m}$$a_t = \frac{mgsin\theta}{m}$$a_t = gsin\theta$ from above equation$sin\theta = \frac{\sqrt{5}}{3}$so tangential acceleration will be $a_t = \frac{\sqrt{5}}{3}*g$$a_t = 7.45m/s^2$

one year 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