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shusha [124]
3 years ago
6

A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe

d around the axle, and the top is set spinning by applying T=3.15 N of constant tension to each string. If it takes 0.320 s for the string to unwind, how much angular momentum does the top acquire? Assume that the strings do not slip as the tension is applied.
Physics
1 answer:
tekilochka [14]3 years ago
7 0

Answer:

Angular momentum, L=6.47\times 10^{-3}\ m

Explanation:

It is given that,

Radius of the axle, r=3.21\ mm=3.21\times 10^{-3}\ m

Tension acting on the top, T = 3.15 N

Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

\tau=\dfrac{dL}{dt}

Torque acting on the top is given by :

\tau=F\times r

Here, F is the tension acting on it. Torque acting on the top is given by :

\tau=2F\times r

2T\times r=\dfrac{L}{t}

L=2T\times r \times t

L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32

L=6.47\times 10^{-3}\ m

So, the angular momentum acquired by the top is 6.47\times 10^{-3}\ m. Hence, this is the required solution.

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You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 7.00 m above the c
Mariana [72]

Answer:

Your friend has to wait 0.26 s after you throw the ball to start running.

Explanation:

The equation that gives the position vector of the ball is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal positon

v0 = initial velocity

t = time

α = throwing angle

y0 = initial vertical position

g = acceleration due to gravity

The equation of displacement of your friend is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of your friend at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.

Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:

y = y0 + v0 · t ·sin α + 1/2 · g · t²

-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m

Solving the quadratic equation:

t = 1.71 s

Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):

x= x0 + v0 · t · cos α

x = 0m + 9.00 m/s · 1.71 s · cos 33°

x = 12.9 m

Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.

Let´s see, how much time it takes your friend to run that distance:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0, v0 = 0)

x = 1/2 · a · t²

1.9 m = 1/2 · 1.80 m/s² · t²

Solving for t

t = 1.45 s

Then, since the time of flight of the ball is 1.71 s, your friend has to wait

1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.

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