A pendulum consists of a bob of mass \( m = 0.1 \, \text{kg} \) and a massless, inextensible string of length \( L = 10 \, \text{m} \). It is suspended from a fixed point at a height \( H = 9 \, \text{m} \) above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse \( P = 0.2 \, \text{kg} \cdot \text{m/s} \) is imparted to the bob at some instant.
After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. We are tasked with finding the magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off, and the kinetic energy of the pendulum just after the lift-off.
The given values are:
Since the bob is lifted after the string becomes taut, it will gain potential energy, and we can use the work-energy theorem. The total mechanical energy of the system just after the lift-off will consist of the kinetic energy and potential energy.
From the conservation of mechanical energy and the information provided, the kinetic energy \( K \) of the pendulum just after the lift-off is given by: \[ K = 0.16 \, \text{Joules} \]
The value of \( K \) is \( \boxed{0.16} \, \text{Joules} \).
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
It also means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another. For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes.
So, mathematically we can represent the law of energy conservation as the following,
The amount of energy spent in a work = The amount of Energy gained in the related work
Now, the derivation of the energy conservation formula is as followed,
Ein − Eout = Δ Esys
We know that the net amount of energy which is transferred in or out of any system is mainly seen in the forms of heat (Q), mass (m) or work (W). Hence, on re-arranging the above equation, we get,
Ein − Eout = Q − W
Now, on dividing all the terms into both the sides of the equation by the mass of the system, the equation represents the law of conservation of energy on a unit mass basis, such as
Q − W = Δ u
Thus, the conservation of energy formula can be written as follows,
Q – W = dU / dt
Here,
Esys = Energy of the system as a whole
Ein = Incoming energy
Eout = Outgoing energy
E = Energy
Q = Heat
M = Mass
W = Work
T = Time