A 0.300-kg potato is tied to a string with length 2.50 m,...

Name: Problem 53, Chapter 7: University Physics with Modern Physics
Uploaded: 2019-06-29T23:18:23-08:00
Duration: 5 min 48 s
Channel: Ben Nicholson
Description: A 0.300-kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

A 0.300-kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

Key Concepts

Conservation of Mechanical Energy

This concept refers to the principle that, in the absence of non-conservative forces, the total mechanical energy (the sum of potential and kinetic energy) of a system remains constant. In the case of a pendulum released from a height, the gravitational potential energy lost as the object descends is transformed into kinetic energy at the lowest point.

Circular Motion and Centripetal Force

An object moving in a circular path requires a net inward force known as the centripetal force, which keeps it moving along the curved path. For a pendulum at the lowest point, the need for centripetal acceleration results in an additional force requirement that must be provided by the tension in the string.

Tension Analysis in Pendular Motion

The tension in a pendulum string not only supports the weight of the suspended mass but also supplies the centripetal force required for circular motion at the lowest point. This leads to a combined force that is greater than the mere gravitational force, calculated typically by adding the weight to the mass times the square of the speed divided by the string length.

Transcript

00:01 Problem 7 .53.

00:03 For some reason we've tied a potato to the end of the string and we've now fastened the other end of the string to see it out horizontally the string top and we drop it from rest and we want to know at the bottom of the swing it goes through what its speed will be and with the tension in our string so let's use conservation of energy and the string is, as a potato and the string, is initially addressed, so there's only going to be potential energy from gravity.

00:47 No work is being done on the string by anything other than gravity.

00:51 The tension in the string is always perpendicular to its direction of motion, so it can't do any work.

01:02 And then this is going to be equal to, if we call this point, the zero point for gravitational energy, then its potential energy will be zero and it will be all kinetic energy.

01:17 So our masses cancel out and our final speed is going to be two times acceleration of gravity times the height change in height that's just going to be our two and a half meters because it's only the height that matters not the x component of displacement because gravity is only pointing up and down and on either side.

02:05 So this works out to be 7 meters per second.

02:22 So for part b now, let's draw a free body diagram.

02:42 We have the tension in the string.

02:47 We have the weight of the potato.

02:50 Now because this is moving in a circle, it has to have a centripetal for it.

02:57 Course, look at the m, ac is b squared over r...

Please give Ace some feedback