Manish Arya Physics Tutor [ AP/IIT-JEE/NEET]
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I love to teach students online. It's really an AHA moment for me when I break down the complex concepts to be understood by my students with ease and comfort.

My style of teaching involves starting with the simplest concept all the way up to a mixture of concepts along with examples and intuitions.

My objective of teaching is to make students feel so much so that they are able to solve problems on their own.

Last but not the least : "My happiness derived from teaching is directly proportional to the understanding made by my students."

Here is a sample question and how I explain:

A uniform rod of length L, lying in vertical plane, is hinged at one of its end. The hinge is assumed to be frictionless. The rod is free to rotate about the horizontal axis passing through the hinge in the vertical plane. At the initial moment, rod is held horizontal. Now, the rod is released. Find angular speed of the rod when rod becomes vertical (rod rotates by 90 degree).

Solution:
Following are the steps to solve this problem:

Step 1:
Rod is in pure rotation about the horizontal axis passing through one of its end. So moment of inertia of the rod about axis passing through one of its end, I = (1/3)*M*L^2.
where M and L are the mass and length of the rod.

Step 2:
Kinetic energy of the rigid body is given by KE = (1/2)*I*w^2 where I is the moment of inertia. w is the angular speed of the rigid body. So, kinetic energy of the rod when it becomes vertical is, K = (1/2)*((1/3)*M*L^2)*w^2. (Plugging the value of I from step 1 into KE equation). w is the angular speed of the rod when it becomes vertical.

Step 3:
When a particle is lowered by a vertical distance of h. Its change in gravitational potential energy is given by ΔU = -m*g*h. Negative sign indicates potential energy is decreasing.
When rod becomes vertical, change in its gravitational potential energy is
ΔU = -M*g*(L/2). Here, we have taken L/2 because centre of mass of the rod falls by a vertical distance of L/2.

Step 4:
Apply law of conservation of energy. As hinge is frictionless, no heat is lost. so,
ΔKE + ΔU = 0 ----- (1)
Initial kinetic energy of the rod is zero as it is released from rest. Final KE when rod becomes vertical is given by K in step 2. Value of ΔU is plugged from step 3.

So equation (1) follows

K-0 + (- M*g*(L/2)) = 0 where K = (1/2)*((1/3)*M*L^2)*w^2

or, (1/2)*((1/3)*M*L^2)*w^2 + (- M*g*(L/2)) = 0
or, (1/3)*L*w^2 - g = 0 ( taking M*(L/2) common )
or, w^2 = 3*g/L

hence, w = √ (3*g/L) (√ is a square root symbol)

Subjects

  • Maths Grade 11-Grade 12

  • Physics Grade 11-Bachelors/Undergraduate

  • Python Beginner-Expert

  • Machine Learning Beginner-Expert

  • Multivariable Calculus Beginner-Expert

Experience

  • Associate Professor (Jun, 2011–Jul, 2014) at FIITJEE LTD
    Physics Faculty for IIT aspirants

Education

  • B. Tech (May, 2005–Mar, 2009) from NIT Jamshedpur, Jamshedpur–scored 9.2

Fee details

    ₹1,500/hour (US$17.96/hour)

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