A cat walks along a plank with mass M= 7.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
1 answer:
Answer:
Before the plank will tip cat will walk 1.652 m
Explanation:
Mass of the cat along with plank
Center of mass of the plank
Mass of cat
We have to find how far right of sawhorse B.
Plank will tip when weight of the cat about B is greater than the torque by the weight of the plank.
Balancing the the torque
You might be interested in
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 1) : 62.5 km/hour is the average velocity of the train.
2) The final velocity of the car at the end of 75 m is 14.69 m/s
Explanation:
1) Displacement of the train = 100 km + 150 km = 250 km
Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours
Average velocity=
62.5 km/hour is the average velocity of the train.
2) The acceleration of the car, a= 1.2
Distance covered by the car,s = 75 m
Initial velocity of the car , = 6 m/s
Final velocity of thre car ,=?
Using third equation of motion:
The final velocity of the car at the end of 75 m is 14.69 m/s