A boy attaches a 0.5kg rock to a 1.3m long massless cord and twirls the rock in a circle over his head so that the circle is parallel to the ground. The stone is moviung in a circle with constant speed. The angle the cord makes with the horizontal is 15 degrees.
a) What is the radius of the circe that the stone makes as it is twirled around?
I got 1.256m, is that right?
b) What is the centripital force on the stone?
I got .5v^2/1.256 but I don't know how to find v.. help!
c) What is the speed of the stone as it moves in the circle?
I know it will just be the magnitude of v... if only I could find v from the earlier section!
d) how much work is gravity performing on the stone?
I have no idea how to go about that one...
a) What is the radius of the circe that the stone makes as it is twirled around?
I got 1.256m, is that right?
b) What is the centripital force on the stone?
I got .5v^2/1.256 but I don't know how to find v.. help!
c) What is the speed of the stone as it moves in the circle?
I know it will just be the magnitude of v... if only I could find v from the earlier section!
d) how much work is gravity performing on the stone?
I have no idea how to go about that one...
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a) Yes. But give the answer to 3 significant figures, you can't justify 4. (But use the more precise value in the later part of the question).
b) The weight of the rock = mg = 0.5x9.8 N
In a vertical direction, the vertical component of tension (T) is equal to the weight:
Tsin(15º) = 0.5x9.8
So you can find T.
The centripetal force (CF) is the horizontal component of T, i.e CF = Tcos(15º)
c) Now you can use mv²/r = CF to find v.
d) Zero. Work = force x distance moved in direction of force. Gravity acts vertically downwards and the stone is not moving up or down, so distance moved in direction of force = 0.
b) The weight of the rock = mg = 0.5x9.8 N
In a vertical direction, the vertical component of tension (T) is equal to the weight:
Tsin(15º) = 0.5x9.8
So you can find T.
The centripetal force (CF) is the horizontal component of T, i.e CF = Tcos(15º)
c) Now you can use mv²/r = CF to find v.
d) Zero. Work = force x distance moved in direction of force. Gravity acts vertically downwards and the stone is not moving up or down, so distance moved in direction of force = 0.
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The angle of 15° that the cord takes with the horizontal forms the necessary force component of tension in the cord to balance the weight of the rock. In other words, if T = tension in cord, then:
T(sin 15°) = mg = (0.5)(9.8)
T = 4.9/sin 15° = 19 N
The horizontal component of T supplies the Centripetal force
required to keep rock at constant speed in circular motion...so
Tcos 15° = mV²/R
19(cos 15) = 18.4 N = (0.5)V²/1.256 {one equation and one unknown}
U can now solve for V :>)
T(sin 15°) = mg = (0.5)(9.8)
T = 4.9/sin 15° = 19 N
The horizontal component of T supplies the Centripetal force
required to keep rock at constant speed in circular motion...so
Tcos 15° = mV²/R
19(cos 15) = 18.4 N = (0.5)V²/1.256 {one equation and one unknown}
U can now solve for V :>)