Two objects of same material are travelling near you. Object A is a 1.1 kg mass traveling 10.2 m/s; object B is a 2 kg mass trav
1 answer:
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Answer:
acceleration are
hollow cylinder < hollow sphere < solid cylinder < solid sphere
Explanation:
To answer this question, let's analyze the problem. Let's use conservation of energy
Starting point. Highest point
Em₀ = U = m g h
Final point. To get off the ramp
Em_f = K = ½ mv² + ½ I w²
notice that we include the kinetic energy of translation and rotation
energy is conserved
Em₀ = Em_f
mgh = ½ m v² +1/2 I w²
angular and linear velocity are related
v = w r
w = v / r
we substitute
mg h = ½ v² (m + I / r²)
v² = 2 gh
v² = 2gh
this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)
v² = v₀² + 2 a L
where L is the length of the plane
v² = 2 a L
a = v² / 2L
we substitute
a =
let's use trigonometry
sin θ = h / L
we substitute
a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }
the moment of inertia of each object is tabulated, let's find the acceleration of each object
a) Hollow cylinder
I = m r²
we look for the acerleracion
a₁ = g sin θ 1/1 + mr² / mr² =
a₁ = g sin θ ½
b) solid cylinder
I = ½ m r²
a₂ = g sin θ = g sin θ
a₂ = g sin θ ⅔
c) hollow sphere
I = 2/3 m r²
a₃ = g sin θ
a₃ = g sin θ
d) solid sphere
I = 2/5 m r²
a₄ = g sin θ
a₄ = g sin θ
We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)
a) a₁ = g sin θ ½ = g sin θ
b) a₂ = g sinθ ⅔ = g sin θ
c) a₃ = g sin θ = g sin θ
d) a₄ = g sin θ = g sin θ
the order of acceleration from lower to higher is
a₁ <a₃ <a₂ <a₄
acceleration are
hollow cylinder < hollow sphere < solid cylinder < solid sphere
Answer:
a
The radial acceleration is
b
The horizontal Tension is
The vertical Tension is
Explanation:
The diagram illustrating this is shown on the first uploaded
From the question we are told that
The length of the string is
The mass of the bob is
The angle made by the string is
The centripetal force acting on the bob is mathematically represented as
Now From the diagram we see that this force is equivalent to
where T is the tension on the rope and v is the linear velocity
So
Now the downward normal force acting on the bob is mathematically represented as
So
=>
=>
The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as
=>
substituting values
The horizontal component is mathematically represented as
substituting value
The vertical component of tension is
substituting value
The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is
substituting value
