Q.A motorcyclist (as a particle) is undergoing vertical circles inside a sphere of death. The speed of the motorcycle varies between 6 m/s and 10 m/s. Calculate the diameter of the sphere of death. What are the minimum values possible for these two speeds?

From $v_{bottom}^2-v_{top}^2=4gr$ with the given 10 and 6 m/s, $r=1.6$ m so diameter = 3.2 m; the theoretical minimum speeds at top/bottom for this radius are $(v_1)_{min}=\sqrt{rg}=4$ m/s and $(v_2)_{min}=\sqrt{5rg}=4\sqrt5$ …

The given operating speeds, 6 m/s (at the top, the slower extreme) and 10 m/s (at the bottom, the faster extreme), are related by energy conservation exactly as in Eq. (1.12): $v_{bottom}^2-v_{top}^2=4gr \;\Rightarrow\; (10)^2-(6)^2=4(10)r \;\Rightarrow\; 100-36=40r \;\Rightarrow\; r=\frac{64}{40}=1.6\text{ m}$ so the sphere's diameter is $2r=3.2$ m.

With this radius now known, the THEORETICAL minimum possible speeds at the top and bottom (the bare minimum needed for the normal reaction N to stay non-negative, i.e. for the motorcyclist to just maintain contact) are, from Eqs. (1.10) and (1.13): $(v_1)_{min}=v_{A,min}=\sqrt{rg}=\sqrt{1.6\times10}=\sqrt{16}=4\text{ m/s}$ $(v_2)_{min}=v_{B,min}=\sqrt{5rg}=\sqrt{5\times16}=\sqrt{80}=4\sqrt5\approx8.94\text{ m/s}$ Since the …

Use energy conservation ($v_{bottom}^2-v_{top}^2=4gr$) between the given actual top and bottom speeds to find the radius, then substitute that radius into the theoretical minimum-speed formulas $\sqrt{rg …

Assuming the GIVEN speeds (6 and 10 m/s) already ARE the theoretical minimum speeds, rather than realising the question separately asks for what the theoretical minimums would be (which turn out to …