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?
Both spheres are identical.
∴ Total energy,
TE = KE + PE
KE is kinetic energy
PE is potential energy
Since bodies are starting from rest, there initial KE is zero,
TEi = PEi = mgh
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
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
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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A disc of radius R is rotating with an angular speed ω0 about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is μk.
A. What was the velocity of its centre of mass before being brought in contact with the table?
B. What happens to the linear velocity of a point on its rim when placed in contact with the table?
C. What happens to the linear speed of the centre of mass when disc is placed in contact with the table?
D. Which force is responsible for the effects in B. and C.
(e) What condition should be satisfied for rolling to begin?
(f) Calculate the time taken for the rolling to begin.
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