A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?
Answers
Let mass of each sphere m and radius of sphere is r .and h is height of inclined planes .
If V and w are the linear and angular speed of sphere at bottom of inclined plane , then according to Law of conservation of energy,
Increase in K.E = Decrease in P.E
(1/2)Mv² + (1/2)Iw² = Mgh
For soid sphere, I = (2/5)MR²
angular speed (w) = v/R
(1/2)Mv² + 1/2× (2/5)MR² ×v²/R² = Mgh
(7/10)Mv² = Mgh
v = √{10gh/7}
Speed is independent of mass, inclined plane so, speed on each cases are same .
(B) yep, the sphere will take longer time to roll down on plane with smaller angle of inclination.
(C) Let t1 and t2 time taken by the sphere to roll down on each plane .
We know,
a = gsin∅/(1 + k²/R²)
Where K is radius of gyration.
For solid sphere,
K = (2/5)R²
Acceleration for 1st inclined plane (a1) = gsin∅1/(1 + 2/5)
= (5/7)gsin∅1
And 2nd inclined plane (a2) = 5/7gsin∅2
We also know,
V = U + at
Initial speed = 0
Final speed in each cases are same .
So, at will be constant
Hence, t inversely proportional to a .
t1/t2 = a2/a1 = sin∅2/sin∅1
If ∅2 > ∅1
Then,
t1 > t2
We observed that ,
The sphere will take longer time to roll down the plane with smaller angle of inclination
(a) Mass of the sphere = m
Height of the plane = h
Velocity of the sphere at the bottom of the plane = v
At the top of the plane, the total energy of the sphere = Potential energy = mgh
At the bottom of the plane, the sphere has both translational and rotational kinetic energies.
Hence, total energy = (1/2)mv2 + (1/2) I ω2
Using the law of conservation of energy, we can write:
(1/2)mv2 + (1/2) I ω2 = mgh
For a solid sphere, the moment of inertia about its centre, I = (2/5)mr2
Hence, equation (i) becomes:
(1/2)mr2 + (1/2) [(2/5)mr2]ω2 = mgh
(1/2)v2 + (1/5)r2ω2 = gh
But we have the relation, v = rω
∴ (1/2)v2 + (1/5)v2 = gh
v2(7/10) = gh
v = √(10/7)gh
Hence, the velocity of the sphere at the bottom depends only on height (h) and acceleration due to gravity (g). Both these values are constants. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.
(b) Consider two inclined planes with inclinations θ1 and θ2, related as:
θ1 < θ2
The acceleration produced in the sphere when it rolls down the plane inclined at θ1 is:
g sin θ1
R1 is the normal reaction to the sphere.
Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at θ2 is:
g sin θ2
R2 is the normal reaction to the sphere.
θ2 > θ1; sin θ2 > sin θ1 … (i)
∴ a2 > a1 … (ii)
Initial velocity, u = 0
Final velocity, v = Constant
Using the first equation of motion, we can obtain the time of roll as:
v = u + at
∴ t ∝ (1/α)
For inclination θ1 : t1 ∝ (1/α1)
For inclination θ2 : t2 ∝ (1/α2)
From above equations, we get:
t2 < t1
Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.