Category: HC Verma Part 1: Mechanics
-
HC Verma Chapter 7 Problem 5 — Centripetal force on car taking a turn
Problem Statement Solve the Newton’s Laws / mechanics problem: Solve the Newton’s Laws / mechanics problem: A 1200 kg car takes a circular turn of radius 50 m at 15 m/s. Find the centripetal force. See problem statement for all given quantities. Circular motion requires a centripetal force directed toward the centre, providing the centripetal…
-
HC Verma Chapter 7 Problem 3 — Speed of a point on Earth’s equator
Problem Statement Solve the kinematics problem: Find the linear speed of a point on the Earth’s equator due to Earth’s rotation. ($R_E = 6400$ km) All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) This problem draws on fundamental physical principles. The key is to iden…
-
HC Verma Chapter 7 Problem 4 — Centripetal acceleration at equator
Problem Statement Solve the Newton’s Laws / mechanics problem: Solve the Newton’s Laws / mechanics problem: Find the centripetal acceleration of a point on the Earth’s equator. ($R_E = 6.4\times10^6$ m, $\omega = 7.27\times10^{-5}$ rad/s) See problem statement for all given quantities. Circular motion requires a centripetal force directed toward the centre, pro Given Information…
-
HC Verma Chapter 7 Problem 2 — Angular velocity of minute hand of clock
Problem Statement Solve the rotational mechanics problem: Solve the kinematics problem: Find the angular velocity of the minute hand of a clock. See problem statement for all given quantities. Rotational kinematics mirrors linear kinematics with $\theta \leftrightarrow x$, $\omega \leftrightarrow v$, $\alpha \leftrightarrow a$. The angular velocity vector Given Information Mass $m$, geometry (radius $R$,…
-
HC Verma Chapter 7 Problem 1 — Angular velocity of Earth
Problem Statement Solve the rotational mechanics problem: Solve the kinematics problem: Find the angular velocity of the Earth’s rotation about its own axis. See problem statement for all given quantities. Rotational kinematics mirrors linear kinematics with $\theta \leftrightarrow x$, $\omega \leftrightarrow v$, $\alpha \leftrightarrow a$. The angular vel Given Information Mass $m$, geometry (radius $R$,…
-
HC Verma Chapter 6 Problem 40 — Car taking turn: maximum speed without friction
Problem Statement Solve the Newton’s Laws / mechanics problem: A car takes a turn of radius 80 m on a flat road with $\mu_s = 0.6$. Find the maximum speed without skidding. ($g = 10$ m/s²) Centripetal force ≤ max friction: $mv^2/r \leq \mu_s mg$ → $v \leq \sqrt{\mu_s rg}$ Step 1: $v_{max} = \sqrt{\mu_s…
-
HC Verma Chapter 6 Problem 39 — Tension in three-block system with friction
Problem Statement Solve the Newton’s Laws / mechanics problem: Three blocks of masses 2, 3, 4 kg are on a rough surface ($\mu_k = 0.2$) and connected by strings. A 30 N force pulls the 4 kg block. Find the acceleration and tensions. ($g = 10$ m/s²) $F – f_{k,total} = (m_1+m_2+m_3)a$ Step 1: $f_k…
-
HC Verma Chapter 6 Problem 38 — Coefficients from two experiments on incline
Problem Statement Solve the Newton’s Laws / mechanics problem: A block placed on a 30° incline stays at rest. When the angle is increased to 45°, it slides. Given $\cos30°=0.866$, $\cos45°=0.707$. Find the range of $\mu_s$. $\mu_s \geq \tan30° = 0.577$ (stays at rest at 30°); $\mu_s Step 1: At 30°: $\mu_s \geq \tan30° =…
-
HC Verma Chapter 6 Problem 36 — Minimum μ for block not to slide on accelerating truck
Problem Statement A block sits on a truck accelerating at 4 m/s². Find the minimum $\mu_s$ between block and truck to prevent sliding. ($g = 10$ m/s²) Given Information All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) Physical Concepts & Formulas This problem draws on…
-
HC Verma Chapter 6 Problem 37 — Block on incline: find time to slide down
Problem Statement Solve the Newton’s Laws / mechanics problem: A 2 kg block is released from rest on a 45° rough incline of length 5 m ($\mu_k = 0.2$). Find the time to reach the bottom. ($g = 10$ m/s²) $a = g(\sin45° – \mu_k\cos45°)$; $s = \frac{1}{2}at^2$ Step 1: $a = 10 \times \frac{1}{\sqrt{2}}(1-0.2)…