Angular Acceleration and Projectile Motion

May 2009
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An amusement park game, shown in the figure (see attachment), launches a marble toward a small cup. The marble is placed directly on top of a spring-loaded wheel and held with a clamp. When released, the wheel spins around clockwise at constant angular acceleration, opening the clamp and releasing the marble after making 11/12 revolution.

What angular acceleration is needed for the ball to land in the cup?
 

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Parvez

PHF Hall of Honor
Feb 2009
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India
Please go through the attachment.
The marble is released after 11/12 revolution. 11/12 revolution = 360*11/12 = 330 degree. It means marble is released from a point making an angle 360-330 = 30 degree with the vertical diameter and on the left of the diameter. This also means that the marble is projected making an angle of 30 degree with the horizontal (upward) from a height rcos 30 = 0.2*(sqrt3)/2 m and has a range equal to (1 + rsin30)m = 1+0.2*1/2 = 1.1 m.
From this data you can find velocity of projection v(say). Angular velocity w' at the time of release will therefore be v/r. (r is the radius).Initial angular velocity w is zero and the angular displacement @ is equal to 2II*11/12 = 2*3.14*11/12 rad. Use this data in equation \(\displaystyle w'^2 = w^2 + 2a@ \)to find angular acceleration a.
 
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May 2009
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Please go through the attachment.
The marble is released after 11/12 revolution. 11/12 revolution = 360*11/12 = 330 degree. It means marble is released from a point making an angle 360-330 = 30 degree with the vertical diameter and on the left of the diameter. This also means that the marble is projected making an angle of 30 degree with the horizontal (upward) from a height rcos 30 = 0.2*(sqrt3)/2 m and has a range equal to (1 + rsin30)m = 1+0.2*1/2 = 1.1 m.
From this data you can find velocity of projection v(say). Angular velocity w' at the time of release will therefore be v/r. (r is the radius).Initial angular velocity w is zero and the angular displacement @ is equal to 2II*11/12 = 2*3.14*11/12 rad. Use this data in equation \(\displaystyle w'^2 = w^2 + 2a@ \)to find angular acceleration a.
I'm having trouble finding the velocity of projection.

I know that the x component is: vcos(theta)t = 1.1

However the y component is confusing for me: 0.2*sqrt3/2 + vsin(theta)t - 4.9t^2.

How do I isolate t in the y component? I know I have to use the quadratic formula, but both solutions are viable in this case.
 

Parvez

PHF Hall of Honor
Feb 2009
365
182
India
Imprtant : I am using 3' in place of sqrt3

For horizontal motion you got v(cos@)*t =1.1 => vcos30 =1.1
=> v3'/2=1.1 => t= 2.2/v3' ................(1)

Now consider vertical motion-
(vertical component of the velocity is vsin30 = v/2)
Y = Yo + vt/2 + (1/2)gt^2 ( are you aware of this eq)
0 = 0.2*3'/2 +v(2.2/v3')/2 + (1/2)(-9.8)(2.2/v3')^2. I replaced t for 2.2/v3' and Yo = rcos30 = 0.2*3'/2.
Can you find v from this eq?
 
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May 2009
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Imprtant : I am using 3' in place of sqrt3

For horizontal motion you got v(cos@)*t =1.1 => vcos30 =1.1
=> v3'/2=1.1 => t= 2.2/v3' ................(1)

Now consider vertical motion-
(vertical component of the velocity is vsin30 = v/2)
Y = Yo + vt/2 + (1/2)gt^2 ( are you aware of this eq)
0 = 0.2*3'/2 +v(2.2/v3')/2 + (1/2)(-9.8)(2.2/v3')^2. I replaced t for 2.2/v3' and Yo = rcos30 = 0.2*3'/2.
Can you find v from this eq?
Ah I wasn't aware that we could substitue the x component into the y. Thanks so much!

(Clapping)