The angular velocity of the wheels of a scooter is directly proportional to the petrol input.The scooter is moving on a frictionless road with uniform velocity. If the input is increased by 50%,what is the change in linear velocity?
Ans:zero
pls explain why?
106as there is no friction .. there is no force to provide the torque required to increase or decrease the angular velocity .. even the friction only provides the linear acceleration.. hence even though the petrol input is changed .. there is no increase in either linear or angular velocity...
106adding to richa's statement:
unless there is any initial linear velocity
1ok heres another doubt,
A cubical block of side a slides on a rough inclined plane,the torque due to normal force about COM is 1/2mga.sinθ , why not zero?direcn of normal force passes thru COM
106For equilibrium .. OR motion without rotation of the block there is torque due to frction as well naa? .. so what will oppose it ? the normal reaction will shift its point of action so that it can apply a opposing torque to the frictional torque ..
1@vivek to avoid toppling n balance torque of friction N ll shift n not pass through centre of mass
1one more
3.A closed cylindrical tube containing some water(not filling the entire tube),lies in the horizontal plane . If it is rotated about a perpendicular bisector,The MOI of water about the axis
A)increases
B)decreases
C)remains constant
D) increases if rotation is clockwise,decreases otherwise
62Vivek.. it does not..
solve this simple(?) question
A block is on the ground with coeff of friction k
a horizontal force F is acting on it
the block is of dimension such that it is a cube of side a
friction is large enuf to prevent the block from moving.
Find the normal acting ? and the point where it acts
113 increases
as water tends to move outwards the effective radius increases
1does it act ka/2 parallel to the centre?
(when f is max,otherwise less than that)
106if there is sufficient friction to keep the block at rest, and friction is not limiting ... then magnitude of frictional force = mgsinθ
so torque of friction abt COM = a/2*mgsinθ
torque of N abt COM should be equal and opposite to the torque by friction abt COM.. so torque by N = mgasinθ/2
62yes asish.. good work
vivek do you now understand why the normal reaction is not always thru the CM of the block?