Minimum Height for Loop Completion

To successfully complete the circular loop, the object must maintain contact with the track even at the highest point. This requires a minimum speed at the top of the loop.

Condition at the Top of the Loop

Let the radius of the circular loop be $R$. The diameter is given as $D$, so $R = D/2$. The object needs enough speed $v_{top}$ at the top of the loop to provide the necessary centripetal force ($m v_{top}^2 / R$). At the minimum condition for completing the loop, the normal force ($N$) exerted by the track becomes zero, and only gravity provides the centripetal force.

• Forces at the top: Gravity ($mg$) downwards, Normal force ($N$) downwards.
• Centripetal force equation: $N + mg = \frac{m v_{top}^2}{R}$
• Minimum condition ($N=0$): $mg = \frac{m v_{top}^2}{R}$
• This simplifies to: $v_{top}^2 = gR$

Energy Conservation

The object starts from rest at an initial height $h$ (the height of the inclined track) and reaches the top of the loop (height $2R$ or $D$) with speed $v_{top}$. We can use the principle of conservation of mechanical energy, as the track is frictionless.

• Initial Energy (at height $h$): $E_{initial} = PE_{initial} + KE_{initial} = mgh + 0 = mgh$
• Energy at the top of the loop (height $2R$): $E_{top} = PE_{top} + KE_{top} = mg(2R) + \frac{1}{2} m v_{top}^2$
• Applying conservation of energy ($E_{initial} = E_{top}$): $mgh = mg(2R) + \frac{1}{2} m v_{top}^2$

Calculating Minimum Height

Substitute the condition for minimum speed ($v_{top}^2 = gR$) into the energy conservation equation:

• $mgh = mg(2R) + \frac{1}{2} m (gR)$
• Divide the entire equation by $mg$: $h = 2R + \frac{1}{2} R$
• Combine terms: $h = \frac{4R + R}{2} = \frac{5R}{2}$

Now, substitute $R = D/2$ back into the equation for $h$:

• $h = \frac{5}{2} \left( \frac{D}{2} \right)$
• $h = \frac{5D}{4}$

Therefore, the minimum height of the inclined track required for the object to successfully complete the loop is $\frac{5D}{4}$.