Problem 1.34 — Centripetal acceleration — Earth’s rotation

Problem Statement

Solve the Newton’s Laws / mechanics problem: Find the centripetal acceleration of a point on the Earth’s equator due to Earth’s rotation. Compare with $g$. Data: $R_E = 6.37\times10^6\,\text{m}$, $T = 86400\,\text{s}$ Angular velocity: $\omega = 2\pi/T = 2\pi/86400 = 7.27\times10^{-5}\,\text{rad/s}$ Centripetal acceleration: $$a_c = \omega^2 R

Given Information

• $T = 86400\,\text{s}$

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

$$a_c = \omega^2 R

Given Information

• Mass(es), forces, angles, and coefficients of friction as given
• $g = 9.8\,\text{m/s}^2$ (acceleration due to gravity)

Physical Concepts & Formulas

Newton’s three laws form the complete foundation of classical mechanics. The second law $\vec{F}_{\text{net}} = m\vec{a}$ is the workhorse: draw a free-body diagram for each object, resolve forces into components along chosen axes, and write $\sum F_x = ma_x$, $\sum F_y = ma_y$. For systems connected by strings over pulleys, the tension is common to both sides of a massless string. Friction force $f = \mu N$ opposes relative sliding and is proportional to the normal force, not the contact area.

• $\vec{F}_{\text{net}} = m\vec{a}$ — Newton’s second law
• $f_k = \mu_k N$ — kinetic friction
• $f_{s,\max} = \mu_s N$ — maximum static friction
• $N = mg\cos\theta$ — normal force on incline of angle $\theta$
• $a = g\sin\theta – \mu_k g\cos\theta$ — acceleration on rough incline

Step-by-Step Solution

Step 1 — Free-body diagram: Draw all forces on each object separately.

Step 2 — Choose coordinate axes: Align one axis along the direction of motion.

Step 3 — Apply Newton’s 2nd Law component by component:

$$

$$

Step 4 — Constraint equations: For Atwood machine: $a_1 = -a_2 = a$, same tension $T$.

Step 5 — Solve the system of equations for $a$ and $T$.

Worked Calculation

Substituting all values with units:

Atwood machine: $m_1 = 3\,\text{kg}$, $m_2 = 5\,\text{kg}$:

$$

$$

$$

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.