Calculate the Velocity of a Mass released in a Sphere Cavity

[achiever]

Q6. Inside a fixed sphere of radius R and uniform density d ,there is a spherical cavity of radius R/2 such that the surface of the cavity passes through the centre of the sphere . A particle of mass m is released from rest at centre B of the spherical cavity. Calculate the vel. withwhich the particle strikes the centre A of the Sphere .Neglect Earth's gravity .initially m was at rest

Q1. A mass at C is freely jointed to 2 equal light rods CA and CB .end A is pivoted and B is freely joined to a bead of mass m which slides on a smooth vertical bar with B below A .C rotates in a horizontal circle with uniform w . find the inclination of the rods CA and CB . lenth of either is l.

Q2. A 4 kg block is attached to the ends of a 2m vertical rod by 2 strings of length 1.25m each when the system rotates about theaxis of the rod ,both strings are taut. if the tension in the upper string is 60 N ,find the tension in the lower string and rpm it makes.

Q3. A particle suspended by a string of length R from a fixed point O is given a vel when at the lowest point . the string becomes slack at athe point A which makes an angle 2A with upward vertical through O . The sring becoes instantaneously taut at B which makes an angle 2B which upward vertical through O. with A and B on opposite sides of the vertical through O. find B in terms Please tell which equation to use

Q4. A projectile of mass m is fired from the surface of the Earth at an angle A=60 degrees from the vertical . The initial speed v0 =sqrt.GM/R .how high does the projectile rise .neglect air resistence an Earth's rotation.of A.

Q5. An artificial satellite is moving in circular orbit around the Earth with speed equal to half the magnitude of escape vel. from the earthIf the satellite is stopped suddenly in its orbit and allowed to fall freely on the Earth ,find the speed with which it hits the surface . In this ques what path it will follow .

 

[dx]

Welcome to PF achiever. Please show your attempts.

 

[achiever]

please clear doubt

Q1.

The current vel. of the river grows in proportion to the distance from the bank and reaches the maximum value of v0 in the middle.

v=0 :near banks. A boat is moving along the river in such a manner that it is always perpendicular to the current .The boat in still water is u .Find the distance through which the boat crossing the river will be caried away by the current if the width of the river is c.Also determine the trajectory of the boat.



Q2. A river of width a with straight parallel banks flows due north with u. The points O and A are on the opposite banks and A is due east of O. Co-ordinate axes Ox and Oy are taken in east and north dirn. resp.. A boat ,whose speed is v relative to water ,starts from O and crosses the river .

if the boat is steered due east and u varies with x as :u=x(a-x)v/a^2

find: 1 eqa. of trajectory

2.time taken to cross the river. 3. absolute vel. of boatsman when reaches the opp. bank. 4. disp. of boatsman when he reaches the opp. bank

 

[achiever]

should i write full soln or just tell what i did

 

[dx]

Just show us what you did, and where you got stuck. Then we can help you.

 

[achiever]

initial potential energy = -11GMm/8R where M =d*vol of larger sphere
final energy=-2GmM'/R where M' is mass of hollow sphere.
so ini PE= final PE + 1/2mv^2
but i am not getting correct answer=sqrt(2/3pi G d R^2)

 

[achiever]

Q6
initial potential energy = -11GMm/8R where M =d*vol of larger sphere
final energy=-2GmM'/R where M' is mass of hollow sphere.
so ini PE= final PE + 1/2mv^2
but i am not getting correct answer=sqrt(2/3pi G d R^2)

 

[achiever]

for Q1 i am having doubt that will the angles formed by both rods will be same

 

[achiever]

Q4. conserving energy
-GMm/R +1/2mv^2 = -GMm/R+h where v=v0 cosA
ans. is R/2 but i am not getting that

 

[dx]

Q4. conserving energy
-GMm/R +1/2mv^2 = -GMm/R+h where v=v0 cosA
ans. is R/2 but i am not getting that

The kinetic energy at the maximum height is not zero. The vertical component of v is zero, but it has a non-zero horizontal component.