During the discus event, Ashton swings a discus of mass 2.0 kg in uniform circular motion - Leaving Cert Physics - Question 6 - 2018
Question 6
During the discus event, Ashton swings a discus of mass 2.0 kg in uniform circular motion. The radius of orbit of the discus is 1.2 m and the discus has a velocity o... show full transcript
Worked Solution & Example Answer:During the discus event, Ashton swings a discus of mass 2.0 kg in uniform circular motion - Leaving Cert Physics - Question 6 - 2018
Step 1
(i) Derive an expression to show the relationship between the radius, velocity and angular velocity of an object moving in uniform circular motion.
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Answer
In uniform circular motion, the relationship between the radius (r), tangential velocity (v), and angular velocity (ω) is given by:
ω=rv
This expression shows that angular velocity is the ratio of linear velocity to radius. Thus, angular velocity increases with an increase in velocity or a decrease in radius.
Step 2
(ii) Calculate the angular velocity of the discus immediately prior to its release.
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Answer
Using the formula derived in part (i):
ω=rv=1.2 m20.4 m s−1=17 rad s−1
Therefore, the angular velocity is 17 rad s⁻¹.
Step 3
(iii) Calculate the centripetal force acting on the discus just before Ashton releases it.
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Answer
The centripetal force (F) can be calculated using the formula:
F=mrv2
Substituting the given values:
F=2.0extkg×1.2extm(20.4extms−1)2=693.6extN
Thus, the centripetal force acting on the discus just before release is 693.6 N towards the centre.
Step 4
(i) his velocity in the horizontal direction.
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Answer
To find the horizontal component of the velocity (
v_h), we use:
vh=vcos(θ)
Substituting the known values:
vh=10.9 m s−1cos(43∘)≈7.97 m s−1
Step 5
(ii) the length of the jump.
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Answer
The length of the jump (s) can be calculated using:
s=vt=(vh)(t)=(7.97 m s−1)(1.03 s)≈8.21extm
Step 6
(i) State the principle of conservation of energy.
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Answer
The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transformed from one form to another.
Step 7
(ii) What is meant by the centre of gravity of a body?
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The centre of gravity of a body is the point at which the weight of that body acts, assuming uniform gravitational field conditions.
Step 8
(iii) what is the maximum height above the ground to which he can raise his centre of gravity?
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Answer
Using the conservation of energy principles, the maximum height can be calculated as follows:
mgh=gain in potential energy=loss in kinetic energy
Substituting the relevant values, we set the potential energy gain equal to the kinetic energy lost:
After solving the equations, the maximum height reached by Ashton's centre of gravity is:
h=4.26+0.98=5.24extm
Step 9
(iv) Draw a diagram to show any forces acting on Ashton when he is at his highest point.
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Answer
Draw a free-body diagram showing forces acting on Ashton at his highest point:
Weight acting downward due to gravity
Any additional supporting force, if applicable
Label the downward force as 'Weight', pointing towards the ground.
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