Question

A mass M = 4 kg attached to a string of length L = 2.0 m...

A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle theta θ= 46.5 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?

0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
A mass M = 4 kg attached to a string of length L = 2.0 m...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A mass M = 4 kg attached to a string of length L = 2.0 m...

    A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle 0 = 44.1 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?

  • A mass m = 4.700 kg is suspended from a string of length L = 1.270...

    A mass m = 4.700 kg is suspended from a string of length L = 1.270 m. It revolves in a horizontal circle. The tangential speed of the mass is 2.243 m/s. What is the angle theta between the string and the vertical (in degrees)?

  • A mass m = 4.300 kg is suspended from a string of length L = 1.290...

    A mass m = 4.300 kg is suspended from a string of length L = 1.290 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 3.743 m/s. What is the angle theta between the string and the vertical (in degrees)?

  • A mass m = 7.9 kg is suspended from a string of length L = 1.15...

    A mass m = 7.9 kg is suspended from a string of length L = 1.15 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 2.80 m/s. What is the angle θ between the string and the vertical (in degrees)?

  • 2. [2pt] A mass m = 9.100 kg is suspended from a string of length L...

    2. [2pt] A mass m = 9.100 kg is suspended from a string of length L = 1.210 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 3.089 m/s. What is the angle theta between the string and the vertical (in degrees)? Answer: Submit All Answers

  • This mass (m) on a string (of length L) is moving in a horizontal circle. The...

    This mass (m) on a string (of length L) is moving in a horizontal circle. The string makes an angle of 60 degrees with the vertical. (Express your answers in terms of m, L, g, and theta) Find the tension in the string. Find the speed of the mass.

  • A mass m = 8.3 kg is suspended from a string of length L = 1.33...

    A mass m = 8.3 kg is suspended from a string of length L = 1.33 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 3.27 m/s. What is the angle between the string and the vertical (in degrees)?

  • 4. A small object of mass M is swinging around in a vertical circle held by...

    4. A small object of mass M is swinging around in a vertical circle held by a massless string of length L, attached to a pivot, as shown. Given that the speed of the mass at the bottom of the circle is Vo, find the tension in the string when the mass is at an angle (theta) with respect to the vertical (see figure) Pivot M

  • A rock of mass m = 0.97 kg is attached to a massless string of length...

    A rock of mass m = 0.97 kg is attached to a massless string of length L = 0.71 m. The rock is swung in a vertical circle faster and faster up to a speed of v = 2.4 m/s, at which time the string breaks. What is the magnitude of the tension, in newtons, at which the string breaks?

  • 6) A 0.25 kg mass is hanged from a string with length L = 0.65 m,...

    6) A 0.25 kg mass is hanged from a string with length L = 0.65 m, making a pendulum. The mass is raised so that the taut string makes an angle of θ = 37◦ with the vertical, and then released from rest. When the mass swings past the bottom (where its speed is maximum) it is observed to have a speed of 1.1 m/s. a) Choosing the bottom as your h = 0 level, what is the initial height...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT