A penny slides in a horizontal circle on the inner surface of a hemispherical bo

Question

A penny slides in a horizontal circle on the inner surface of a hemispherical bowl. (“Hemisphere” means exactly half of a hollow sphere.) The radius of the bowl is R but the radius of the horizontal circle that the penny makes is R/2. How fast (in terms of R and g) is the penny moving? (Of course eventually friction will slow the penny down and it will wind up at the bottom of the bowl. But we’re concerned with the situation at the start when it is moving fast in the horizontal circle.)
Suggestion: Draw a free-body diagram for the penny and use “F=ma”. There will be two equations with two unknowns, a normal force and the speed v you are seeking.

Explanation / Answer

unable to draw free body diagram it is showing some error.

but try to visuallise mg is acting downward acceleration is acting inwards. and normal reaction is going to centre of the hemisphere.

N = normal eaction with an angle of 30 degree { tan inverse of ( r/2 / r ) } with verticle line from centre of hemispher.

NCos 30 = mg

and NSin30 = v^2 / (r/2) = 2*v^2/r

v^2= tan30 * mg* r/2

v= sqrt ( mgr /23 )

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.