A circular saw blade with a radius of 0.145 m starts from rest and turns in a vertical plane with a constant angular acceleratio
n of 2.00 rev/s2. After the blade has turned through 155 revs, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor.
How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?
How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?
1 answer:
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The final speed of the orange is 7.35 m/s
Explanation:
The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration towards the ground. So we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time elapsed
For the orange in this problem, we have
u = 0 (it is dropped from rest)
is the acceleration
Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:
Learn more about free fall:
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(a) 296.6 m
The motion of the stone is the motion of a projectile, thrown with a horizontal speed of
and with an initial vertical velocity of
where we have put a negative sign to indicate that the direction is downward.
The vertical position of the stone at time t is given by
(1)
where
h is the initial height
g = -9.81 m/s^2 is the acceleration due to gravity
The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:
y(6.00 s) = 0
Substituting into eq.(1), we can solve to find the initial height of the stone, h:
(b) 176.6 m
The balloon is moving downward with a constant vertical speed of
So the vertical position of the balloon after a time t is
and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:
(c) 198.2 m
In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.
The horizontal speed of the rock is
So the horizontal distance travelled in t = 6.00 s is
Considering also that the vertical height of the balloon after t=6.00 s is
The distance between the balloon and the rock can be found by using Pythagorean theorem:
(di) 15.0 m/s, -58.8 m/s
For an observer at rest in the basket, the rock is moving horizontally with a velocity of
Instead, the vertical velocity of the rock for an observer at rest in the basket is
Substituting time t=6.00 s, we find
(dii) 15.0 m/s, -78.8 m/s
For an observer at rest on the ground, the rock is still moving horizontally with a velocity of
Instead, the vertical velocity of the rock for an observer on the ground is now given by
Substituting time t=6.00 s, we find