Class 10 Physics • Chapter 01 (Deep Detail)
Translational Motion: When a force acts on a rigid body free to move, the body starts moving in a straight line in the direction of force.
Rotational Motion: When a body is pivoted at a point and a force is applied at a suitable point, it rotates about the axis passing through the pivot.
Definition: The turning effect of a force on a body about an axis is called the Moment of Force or Torque.
SI Unit: Newton-metre (Nm) | CGS Unit: Dyne-cm
Relationship: $1 \text{ Nm} = 10^7 \text{ dyne-cm}$
NUMERICAL Calculated the torque produced by a force of 50 N acting at the end of a lever of length 20 cm.
Solution: $\tau = F \times d = 50 \times 0.2 \text{ m} = 10 \text{ Nm}$.
CONCEPTUAL A mechanic can open a nut by applying a force of 150 N while using a lever handle of length 40 cm. How long a handle is required if he wants to open it by applying a force of only 50 N?
Solution:
Moment required $\tau = 150 \times 0.4 = 60 \text{ Nm}$.
New Force $F' = 50 \text{ N}$. Let new length be $d'$.
$50 \times d' = 60 \implies d' = 1.2 \text{ m}$.
CONCEPTUAL A steering wheel of diameter 0.5 m is rotated anticlockwise by applying two forces each of magnitude 5 N. Calculate the moment of the couple applied.
Solution: Moment = Force $\times$ Couple Arm = $5 \times 0.5 = 2.5 \text{ Nm}$.
Two equal and opposite parallel forces, not acting along the same line, form a Couple.
Effect: A couple produces pure rotational motion.
Moment of Couple = Force $\times$ Couple Arm (Distance between forces).
When a number of forces acting on a body produce no change in its state of rest or of motion, the body is said to be in equilibrium.
Static Equilibrium: Body remains at rest under influence of forces. (e.g., Book on a table).
Dynamic Equilibrium: Body remains in motion (constant velocity) under influence of forces. (e.g., Rain drop falling with terminal velocity, Electron revolving around nucleus).
BOARD QUESTION A boy of mass 40 kg sits at a distance of 1.5 m from the fulcrum of a see-saw. Where should another boy of mass 30 kg sit to balance it?
Solution: By Principle of Moments: $40 \times g \times 1.5 = 30 \times g \times d \implies 60 = 30d \implies d = 2 \text{ m}$.
NUMERICAL A uniform metre rule rests horizontally on a knife edge at the 60 cm mark when a mass of 10 g is suspended from one end. Draw a diagram and calculate the mass of the rule.
Solution:
Pivot at 60cm. Mass of rule ($M$) acts at 50cm. Distance = $60-50 = 10 \text{ cm}$.
Mass 10g must be at 100cm (Longer arm) to balance. Distance = $100-60 = 40 \text{ cm}$.
Anticlockwise Moment = Clockwise Moment
$\implies M \times 10 = 10 \times 40 \implies M = 40 \text{ g}$.
According to the Principle of Moments, in equilibrium:
$$ \sum \text{Anticlockwise} = \sum \text{Clockwise} $$(Used to solve measuring/seesaw problems)
Q: If a body is pivoted at its Centre of Gravity, what is the moment of its weight about the pivot?
Ans: Zero. (Because perpendicular distance is zero).
THINKING Where does the centre of gravity of a hollow sphere lie? Is it necessary for the C.G. to be inside the material of the body?
Ans: The CG lies at the geometric centre. No, it is not necessary to be in the material (e.g., Ring, Hollow Sphere).
REASONING Why does a wide-based flask have greater stability?
Ans: A wide base ensures the vertical line through the CG falls within the base even when tilted significantly. A lower CG (due to heavy bottom) also increases stability.
The point about which the algebraic sum of moments of weights of all particles constituting the body is zero. The entire weight of the body can be considered to act at this point.
Note: It is not necessary for the CG to be within the material of the body (e.g., Ring, Hollow Sphere).
REASONING Why does a wide-based flask have greater stability?
Ans: A wide base ensures the vertical line through the CG falls within the base even when tilted significantly. A lower CG (due to heavy bottom) also increases stability.
Motion: When a particle moves with a constant speed in a circular path, its motion is uniform circular motion.
Velocity: Variable (Direction changes continuously). Hence, it is an accelerated motion.
Acceleration: Acts towards the centre (Centripetal Acceleration).
The force acting on a body moving in a circular path, directed towards the centre of the path.
$$ F_c = \frac{mv^2}{r} $$Examples:
A fictitious force (pseudo force) assumed by an observer moving with the body to explain why they feel pushed outwards. It is equal in magnitude to centripetal force but acts away from the centre.
(Real force is only centripetal; centrifugal is a reaction/effect observed in non-inertial frame).
CONCEPTUAL A stone tied to a string is whirled in a horizontal circle. What is the work done by the tension in the string over one complete revolution?
Ans: Zero. The tension (force) acts perpendicular to the displacement at every instant ($90^\circ$). $W = Fs \cos 90^\circ = 0$.
DIFFERENTIATION State one similarity and one difference between Centripetal and Centrifugal force.
Ans:
Similarity: Both have same magnitude ($mv^2/r$).
Difference: Centripetal acts towards centre (Real). Centrifugal acts away from centre (Fictitious).