Problems with velocity/acceleration involving clock arm

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Homework Help Overview

The problem involves analyzing the motion of a traditional watch's second hand, specifically focusing on its speed, velocity, and acceleration over specific time intervals. The subject area encompasses concepts of kinematics and circular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the speed and velocity of the second hand using trigonometric relationships and the cosine law. Some participants question the distinction between speed and velocity, while others emphasize the importance of distance traveled versus displacement.

Discussion Status

The discussion is ongoing, with participants expressing confusion and seeking clarification on the calculations involved. Some guidance has been offered regarding the definitions of speed and velocity, but there is no explicit consensus on the approach to solving the problem.

Contextual Notes

Participants are working under the constraints of the homework assignment, which includes specific time intervals for calculating velocity and acceleration. There is a noted lack of clarity in the original poster's calculations and the need for further assistance.

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Homework Statement


A traditional watch has a second hand 2.0 cm long, from centre to tip

a. what is the speed of the tip of the second hand?
b. What is the velocity of the tip at 20s? 40s? and 60s?
c. What is it's change in velocity between 15s and 30s?
d. What is it's average acceleration durning the same time interval


Homework Equations


d = v * t
sine law
cosine law
SOH CAH TOA

The Attempt at a Solution


a) used SOH CAH TOA to find the third side of the 90 degree (clock) triangle
cos45 = 2/h
cos45h = 2
h = 4cm (rounded)

v = d/t
v = 4cm / 15s
v = 0.3cm/s

therefore the velocity of the clock arm is 0.3cm/s
****attached is the clock and the final 90 degree triangle****

b) I used the cosine law to TRY to find the other side
due to the clock being on 20s the angle from the start to the ending point of 20s is 120 degrees

C^2 = A^2 + b^2 - 2(2)(2)cos120
c^2 = 8 - 4
C^2 = 4
C = 2

this is where I got stuck
****there is an attachment of the triangle I drew****

thanks to anyone willing to help :D
 

Attachments

  • traingle.png
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  • clock.png
    clock.png
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What is the difference between speed and velocity?
 
the only difference I really see would be that a velocity is a vector while speed is a scalar.
 
What is important is the actual distance that the tip of the second hand travels, not the distance between one location and another (which is displacement, not distance traveled).
 
I'm still pretty confused on how to do it :/
 
bump
(come on I need lots of help!)
 
How far does the tip of the second hand travel in 60 seconds?
 

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