Problem Statement
Solve the Newton’s Laws / mechanics problem: A car starts from rest on a circular track of radius $R = 500\,\text{m}$ with tangential acceleration $w_\tau = 0.80\,\text{m/s}^2$. At what time will the normal acceleration equal the tangential? How far has the car gone? Speed at time $t$: $v = w_\tau t$ Normal acceleration: $w_n = v^2/R = w_\tau^
Given Information
- $R = 500\,\text{m}$
- $w_\tau = 0.80\,\text{m/s}$
Physical Concepts & Formulas
Circular motion requires a centripetal force directed toward the centre, providing the centripetal acceleration $a_c = v^2/r = \omega^2 r$. This force is not a new type of force — it is always the resultant of real forces (tension, normal force, friction, gravity) directed inward. At the minimum speed for maintaining contact, the normal force drops to zero.
- $a_c = v^2/R = \omega^2 R$ — centripetal acceleration
- $F_c = mv^2/R$ — net centripetal force needed
- Banked curve: $\tan\theta = v^2/(Rg)$ — ideal banking angle
- Loop minimum speed: $v_{\min} = \sqrt{gR}$ at top (N=0)
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
Full substitution shown in the steps above.
Answer
$$\boxed{a = \dfrac{(m_2-m_1)g}{m_1+m_2}}$$
Physical Interpretation
The centripetal force is not a ‘new’ force but the net inward resultant of real forces. If that resultant falls below $mv^2/r$, the object cannot maintain circular motion and will fly outward — this is the critical condition for minimum speed problems.
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