Some curious students hold a rolling race by rolling four items down a steep hill. The four items are a solid homogeneous sphere
, a thin spherical shell, a solid homogeneous cylinder and a hoop with all its mass concentrated on the hoop's perimeter. All of the objects have the same mass and start from rest. Assume that the objects roll without slipping and that air resistance and rolling resistance are negligible. For each statement below, select True or False.
a) Upon reaching the bottom of the hill, the hoop will have a larger rotational kinetic energy than any of the other objects will when they reach the bottom of the hill.
A: True B: False
b) The hoop reaches the bottom of the hill before the homogeneous cylinder.
A: True B: False
a) Upon reaching the bottom of the hill, the hoop will have a larger rotational kinetic energy than any of the other objects will when they reach the bottom of the hill.
A: True B: False
b) The hoop reaches the bottom of the hill before the homogeneous cylinder.
A: True B: False
1 answer:
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Answer:
Explanation:
The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).
The equation of uniform rectilinear motion (horizontal ) for the x axis is :
x = xi + vx*t Equation (1)
Where:
x: horizontal position in meters (m)
xi: initial horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m/s
The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:
y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)
vfy= v₀y -gt Equation (3)
vfy²= v₀y²-2gH Equation (4)
Where:
y: vertical position in meters (m)
y₀ : initial vertical position in meters (m)
t : time in seconds (s)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
H= hight that reaches the projectile above its starting point (m)
Data
v₀ : total initial speed
θ₀ : angle of v₀ above the horizontal.
g acceleration due to gravity
Calculation of the componentes x-y of the v₀
v₀x = v₀*cos θ₀
v₀y = v₀*sin θ₀
Calculation of the maximum hight that reaches the projectile above its starting point
When the projectile reaches its maximum height (h), vy = 0:
in the Equation (4)
vfy²= v₀y²-2gh
0= v₀y²-2gh
2gh= v₀y²
2gh= (v₀*sin θ₀)²
Answer:
42.69 N and 18.07 N
Explanation:
We are given that
Mass of ladder=6.2 kg
Length of ladder=1.97 m
Distance of Sawhorse A from one end=0.64 m
Distance of sawhorse B from other end=0.17 m
Let center of Ladder=
Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m
Distance of sawhorse B from center of ladder=0.985-0.17=0.815 m
Force one ladder due to gravity=mg=
Where
Torque applied on Sawhorse A=
Torque applied on Sawhorse B=
In equilibrium
Total force=