Question 245: Work, Energy and Power & Circular Motion

Question

A stone is projected at angle $$	heta$$ with velocity $$u$$. If it executes nearly a circular motion at its maximum point for short time, then the radius of the circular path will be ( $$g=$$ acceleration due to gravity)

Accepted Answer

Since, the particle experiences a downward acceleration due to gravity at peak (highest point), and traces a circular arc, it has no component of acceleration in $$x$$-direction.
Let, radius of circle be $$R$$.
Horizontal velocity of the particle of height point.
$$v_x=u\cos 	heta \quad \left(\because a_x=0ight)$$
$$	herefore a=\frac{v_x^2}{R}$$ (for circular motion centripetal acceleration
$$=\frac{v^2}{R}$$)
$$\Rightarrow R=\frac{v_x^2}{a}=\frac{u^2 \cos ^2 	heta}{g}\quad (\because a=g)$$

Answer

$$\frac{u^2}{g}$$

Answer

$$\frac{u^2 \cos ^2 	heta}{g}$$

Answer

$$\frac{u^2 \sin ^2 	heta}{g}$$

Answer

$$\frac{u^2 \cos ^2 	heta}{2 g}$$

Work, Energy and Power & Circular Motion

A stone is projected at angle $\theta with velocity u. If it executes nearly a circular motion at its maximum point for short time, then the radius of the circular path will be ( g=$ acceleration due to gravity)

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