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Gursafal M.
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Physics (Newtonian Mechanics)
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Question:

A roller of mass m and radius r is attached to a spring with spring constant k at the center axis of the roller. The roller is given a velocity v in an upward direction up to an inclined plane which makes an angle 30 degree with the ground (at this moment the spring is at rest). What is the maximum height the roller's center of mass? Given that m=1 kg, r=1 m, v=1 m/s, k=1 N/m and assume g to be 10 m/s^2. (the roller is solid and you can assume pure rolling)

Gursafal M.
Answer:

Assume the maximum height is h and at that moment the elongation of the spring is x. so we get h=x(sin30)=x(0.5) At the maximum height, the roller will not have any linear or angular velocity so it will have zero kinetic energy. At the maximum height, the whole system's energy will contain the potential energy stored in the spring (PE1) and the gravitational potential energy of the roller (PE2). Now we assume the initial level of the center of mass of the roller as the reference level therefore initially the system will have zero potential energy. At the initial moment, the system's energy will contain the kinetic energy of the roller due to linear motion (KE1) and rotational motion (KE2). let the w=angular speed we are told to assume pure rolling therefore v=r*w and w=v/r the moment of inertia of the roller about the axis of rotation(I) = (0.5)mr^2 KE1=(0.5)(mv^2)=0.5 J KE2=(0.5)(0.5)(mr^2)(w^2)=(0.25)*mv^2=0.25 J PE1=(0.5)kx^2=2*kh^2=2h^2 PE2=mgh=10h now initial and final energies are conserved so KE1 + KE2 = PE1 + PE2 After solving this equation for h using the values we calculated we will get: h=0.073 m

Calculus
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Question:

evaluate: int (1/(1+cosx))dx

Gursafal M.
Answer:

multiply numerator and denominator by 1-cosx this gives us int ((1-cosx)/(1-(cosx)^2))dx= int((1-cosx)/(sinx)^2)dx now 1-cosx = 2*(sin(x/2))^2 (using formula for cos2x) sinx can be written as 2(sin(x/2))(cos(x/2)) this makes the equation int (1/(2(cos(x/2))^2))dx = int ((0.5)(sec(x/2))^2)dx as d(tanx)/dx=(secx)^2 therefore d(tan(x/2))/dx=(0.5)(sec(x/2))^2 int ((0.5)(sec(x/2))^2)dx = tan(x/2)

Advanced Physics (Special Relativity)
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Question:

Using Lorentz transformation show that if two events are simultaneous in a reference frame at rest with respect to ground they are not simultaneous in a reference frame moving with respect to the ground at constant velocity

Gursafal M.
Answer:

Consider two events E1 and E2 which are simultaneous in a frame S which is stationary. Now consider a frame S' which is moving in +ve x direction with speed V. Event E1 has co-ordinates (x1,y1,z1,t1) w.r.t S and (x1',y1',z1',t1') w.r.t to S' the same goes for E2. In S E1 and E2 are simultaneous therefore t1=t2. now t1'= k(t1-(vx1)/c^2), t2'= k(t2-(vx2)/c^2) where k=1/sqrt(1-v^2/c^2). now t1'-t2'= -kv(x1-x2)/c^2, the events are not the same therefore x1 and x2 are not same so x1-x2 is not zero therefore t1'-t2' is not zero, proving that the events are not simultaneous in S'

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