A cart of mass $M$ is released from $A$, the highest point of a frictionless track, as shown in the figure. The cart travels along the track and enters the semicircular arc $DBC$ of radius $R$. The heights of the points $A$ and $B$ are $h_1$ and $h_2$ from the ground, respectively. Which of the following quantities does not play any role in ensuring that the cart does not leave the track?
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In gravitational loop-the-loop problems on frictionless tracks, the motion is purely determined by kinematics and energy conservation.
Since both kinetic and potential energy are directly proportional to mass, $M$ cancels out from the equations.
Therefore, the mass of the object never plays a role in loop-the-loop conditions.
Step 1: Understanding the Question:
This question examines the conditions required for a cart to complete a loop-the-loop without losing contact with the track, focusing on which parameters determine this threshold. Step 2: Key Formulas and Approach:
1. At any point along the circular track, the normal force $N$ and gravity provide the centripetal acceleration.
2. At the highest point of the loop (point $B$), the minimum condition to stay on the track is that the normal force $N \ge 0$.
3. Conservation of mechanical energy is used to relate the speeds and heights:
\[ M g h_1 = M g h_2 + \frac{1}{2} M v_B^2 \] Step 3: Detailed Explanation:
Let $v_B$ be the velocity of the cart of mass $M$ at the highest point of the loop $B$.
The forces acting on the cart at $B$ are the gravitational force $Mg$ (downwards) and the normal force $N$ from the track (downwards).
Writing the equation of motion for circular motion at $B$:
\[ N + Mg = \frac{M v_B^2}{R} \implies N = \frac{M v_B^2}{R} - Mg \]
To ensure that the cart does not leave the track, the normal force must be non-negative ($N \ge 0$):
\[ \frac{M v_B^2}{R} - Mg \ge 0 \implies v_B^2 \ge gR \]
Now, we use the conservation of mechanical energy between the starting point $A$ and the highest point $B$:
\[ M g h_1 = M g h_2 + \frac{1}{2} M v_B^2 \]
Dividing the entire equation by the mass $M$:
\[ g h_1 = g h_2 + \frac{1}{2} v_B^2 \implies v_B^2 = 2g(h_1 - h_2) \]
Notice that the mass $M$ completely cancels out of the energy equation.
Substituting this expression for $v_B^2$ into the circular motion condition:
\[ 2g(h_1 - h_2) \ge gR \implies 2(h_1 - h_2) \ge R \]
Since $h_2 = 2R$, we can also write this condition as:
\[ h_1 \ge \frac{5}{2} R \]
The final condition involves $h_1$, $h_2$, and $R$, meaning that these three parameters are vital to determining whether the cart stays on the track.
The mass $M$ does not appear in this final relation and thus plays no role.
Step 4: Final Answer:
The mass $M$ of the cart plays no role in ensuring that the cart does not leave the track, corresponding to Option (A).
