Q. 184.2( 9 Votes )
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?
Answer :
Both spheres are identical.
∴ Total energy,
TE = KE + PE
Where
KE is kinetic energy
PE is potential energy
Since bodies are starting from rest, there initial KE is zero,
TEi = PEi = mgh
Where
m is mass of the bodies
g is acceleration due to gravity
h is height of inclined plane
On reaching bottom, the potential energy drops to zero (PEf = 0)
Therefore, total energy of bodies at bottom are KE of bodies.
TEf = KEf = 1/2 mvf2+ 1/2 I ω2
Where
m is mass of body
vf is velocity attained by body
ω is angular velocity
I is moment of inertia
By Law of conservation of energy, we have,
TEi = TEf
I.e. mgh = 1/2 mvf2+ 1/2 I ω2
= 1/2 mvf2 + 1/2 (2mR2/5) ω2
V = Rω
∴ mgh = 1/2 mvf2+ mvf2/5
vf = (10gh/7)1/2
The final velocity is independent of angle of inclination. Therefore
both bodies will have same final speed on reaching ground.
B & C
When a body is moving on an inclined plane of inclination θ, acceleration acting on body is g (acceleration due to gravity vertically downward)
The acceleration along inclined plane is g sin(θ) component of acceleration.
Let height from which body fall be H.
Then length of inclined plane of inclination θ,
L = H/sin(θ)
We know the equation,
s = u t + 1/2 at2
Where
u is initial velocity
s is distance covered
a is acceleration acting on body
t is time
For case of motion along plane,
u = 0 (since body starts from rest)
s = L = H/sin(θ)
a = g sin(θ)
t is time
Substituting the values,
H/sin(θ) = 0× t + 1/2 g sin(θ) t2
∴ time for reaching ground,
t = (2H/g)1/2 / sin(θ)
From above equation, we can infer that time taken by a body to roll down an inclined plane is inversely proportional to sin of angle of inclination (θ).
Sine increases with increase of angle, thus time for fall decreases for greater inclination (t ∝ 1/sin(θ))
Thus greater the angle of inclination, lower the time needed for reaching ground.
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