How far from the point of release will the ball hit the ground?

[Mohammad Ishmas]

Homework Statement

Mitchell Starc is bowling to Virat Kohli during the Border-Gavaskar trophy. Just before the release, his arm swings downward like a pendulum with an increasing velocity.
He starts with an angular velocity of 40 rad/s and increases it to 80 rad/s in just 1 second, maintaing roughly circular motion with an effective arm length of 1 m . At the moment of release , he lets go of the ball at an angle of 30 degree from the vertical. The point of release is at a height of 2 m from the ground .

Assuming the pitch is 20 m long , the mass of the ball is 0.15 kg, and g = 10 m/s2 and neglecting air resistance
Then Find how far rom the point of release will the ball pitch or hit the ground

Relevant Equations

Equations of motions and relation between angular and linear variables

I have found an answer in terms of x and y components of the linear velocity of the ball , but I don't know how to relate linear velocity with angular velocity . I know there is a formula regarding it but I want to know the derivation behind it

 

[haruspex]

(Assuming uniform acceleration, the average rotation rate over that second is 60 rad/s, implying the arm rotated through 60 radians.)

The definition of a radian is such that if an arm of length r rotates through one radian then the fist has moved along an arc length r. You can work it out from there.

 

[PeroK]

Homework Statement:
He starts with an angular velocity of 40 rad/s and increases it to 80 rad/s in just 1 second, maintaing roughly circular motion with an effective arm length of 1 m . At the moment of release , he lets go of the ball at an angle of 30 degree from the vertical. The point of release is at a height of 2 m from the ground .

If you accelerate from 40 rad/s to 80 rad/s in 1 second, then you rotate about 60 radians, which is about 10 revolutions. I've never seen bowling like that.

 

[kuruman]

If you accelerate from 40 rad/s to 80 rad/s in 1 second, then you rotate about 60 radians, which is about 10 revolutions. I've never seen bowling like that.

Then you have never seen Mitchell Starc bowl for the Border-Gavaskar trophy.

 

[Mohammad Ishmas]

(Assuming uniform acceleration, the average rotation rate over that second is 60 rad/s, implying the arm rotated through 60 radians.)

The definition of a radian is such that if an arm of length r rotates through one radian then the fist has moved along an arc length r. You can work it out from there.

I think I got you, you are guiding me towards the arc length formula that is , s = r/θ . How does this help with relation between linear velocity and angular velocity ?

 

[Mohammad Ishmas]

I think I got you, you are guiding me towards the arc length formula that is , s = r/θ . How does this help with relation between linear velocity and angular velocity ?

Ohh , I got it . We can just divide both sides by time taken and then we'll obtain the relation , thank you .

 

[haruspex]

I think I got you, you are guiding me towards the arc length formula that is , s = r/θ . How does this help with relation between linear velocity and angular velocity ?

That would mean rotating twice as far would traverse half the arc length. Think again.

 

[erobz]

Assuming the pitch is 20 m long , the mass of the ball is 0.15 kg, and g = 10 m/s2 and neglecting air resistance
Then Find how far rom the point of release will the ball pitch or hit the ground

Should this make sense to me and I'm missing something?

 

[haruspex]

Should this make sense to me and I'm missing something?

Could be two meanings of "pitch": as a noun, an area of turf; as a verb, to land.

 

[PeroK]

Then you have never seen Mitchell Starc bowl for the Border-Gavaskar trophy.

All the data in the problem statement are nonsensical.

 

[Mohammad Ishmas]

All the data in the problem statement are nonsensical.

This problem was given by my physics teacher. I have confirmed that this is the data only , maybe he just gave the data just for the sake of the problem.

 

[Mohammad Ishmas]

That would mean rotating twice as far would traverse half the arc length. Think again.

oh sorry , I mistyped the formula , it is actually s = rϴ, now we can divide both side by time and then get the relation , linear velocity (v) = rω, here ω is the angular velocity

 

[PeroK]

This problem was given by my physics teacher. I have confirmed that this is the data only , maybe he just gave the data just for the sake of the problem.

It's not hard to look up Starc's bowling speed, which is about 40m/s. With a 1m arm, that's not compatible with arm rotation of 80 rad/s, which implies a release speed of about 80m/s. Moreover, with a 1m arm and a release 2m off the ground means his shoulder is about 1m off the ground!

Why not put in a bit of research and set a realistic question?

 

[haruspex]

oh sorry , I mistyped the formula , it is actually s = rϴ, now we can divide both side by time and then get the relation , linear velocity (v) = rω, here ω is the angular velocity

Right.
There are a few ambiguities in the question. What angular velocity is implied at the moment of release? Presumably the 80 rad/s.
Is the release 30 degrees before the vertical or after the vertical? After, I assume.
The point of release is given as 2m above the ground, and it asks how far from that point the ball hits the ground. But likely it just means the horizontal distance.

 

[Mohammad Ishmas]

Right.
There are a few ambiguities in the question. What angular velocity is implied at the moment of release? Presumably the 80 rad/s.
Is the release 30 degrees before the vertical or after the vertical? After, I assume.
The point of release is given as 2m above the ground, and it asks how far from that point the ball hits the ground. But likely it just means the horizontal distance.

I think the release of 30 degrees is after the vertical such that it makes a acute angle with the horizontal , this is assuming the natural bowling action of any bowler , but physically there can be two cases. And yeah the question is asking horizontal distance only.

 

[jbriggs444]

It's not hard to look up Starc's bowling speed, which is about 40m/s. With a 1m arm, that's not compatible with arm rotation of 80 rad/s, which implies a release speed of about 80m/s. Moreover, with a 1m arm and a release 2m off the ground means his shoulder is about 1m off the ground!

Video evidence shows that he uses an overhand throw with a release angle below the horizontal so that the ball travels on a downward trajectory, striking the ground perhaps one or two meters before reaching the batter. The point of release is indeed close to 2m above the ground.

He is in a massive forward lean at the point of release. At release, his shoulder is plausibly at 1 m off the ground. However, the angle of the arm and the angle of the ball's trajectory at the moment of release may not be at right angles to one another. One gets more pitch speed from a flexible human arm than from a rigid bar.

If there is a "downward like a pendulum" motion, it must be an inverted pendulum. If he starts at 40 rad/s, ends at 80 rad/s and does this over a downward circular arc covering only 30 degrees (0.52 rad) at an average rotation rate of 60 rad/s, that would take only about 8.7 milliseconds, not one second.

If the initial velocity were actually 30 degrees below the horizontal, we would expect an impact point just about 4 meters in front of Starc and a bounce taking the ball well over the batter's head.

Why not put in a bit of research and set a realistic question?

Indeed.

 

[PeroK]

Video evidence shows that he uses an overhand throw with a release angle below the horizontal so that the ball travels on a downward trajectory, striking the ground perhaps one or two meters before reaching the batter. The point of release is indeed close to 2m above the ground.

He is in a massive forward lean at the point of release. At release, his shoulder is plausibly at 1 m off the ground. However, the angle of the arm and the angle of the ball's trajectory at the moment of release may not be at right angles to one another. One gets more pitch speed from a flexible human arm than from a rigid bar.

If there is a "downward like a pendulum" motion, it must be an inverted pendulum. If he starts at 40 rad/s, ends at 80 rad/s and does this over an downward circular arc covering only 30 degrees (0.52 rad) at an average rotation rate of 60 rad/s, that would take only about 8.7 milliseconds, not one second.

If the initial velocity were actually 30 degrees below the horizontal, we would expect an impact point just about 4 meters in front of Starc and a bounce taking the ball well over the batter's head.


Indeed.

The massive forward lean is after the ball has been released. The classic fast bowler action is to have a high release point. Starc is nearly 2m tall, so the release point must be over 2m.

If you want to see something unorthodox look up Lasith Malinga of Sri Lanka.

A fast bowler typically has three points on the pitch to aim at. The first is "yorker" length, where he's aiming at the batsman's feet. Trying to get the ball to pitch under the bat. Or break the batsman's big toe!

A "good" length is a few metres in front of the batsman. That's the most common length.

Short-pitched bowling involves pitching the ball half way down the pitch and getting the ball to bounce head high. These deliveries are often called "bouncers" and characterise the aggressive fast bowler.

A ball that pitched only 4m down the pitch would be classed as a "long hop" and seen as an easy one for the batsman to hit.

Note also that the bowler is not allowed to flex the arm during the bowling motion. I.e. the ball must be "bowled" with a straight arm and not thrown. A completely straight arm is virtually impossible, but bowlers whose action is suspect are referred to video analysis and may be banned until they have corrected their action.