Comp Sci

English

Language

Math

Science

Social Science

Test Prep

A bug is on the rim of a disk of diameter 15 in that moves from rest to an angular speed of 71 rev/min in 5.2 s.

1. What is the tangential acceleration?

2.What is the tangential velocity of the bug at the end of the 5.2 s?

3. One second after the bug starts from rest, what is its total acceleration?

4. What angle does this acceleration make with ac?

1 Answers

Hello! Thank you for the question.

The first thing we need to find for this problem is the angular velocity of the bug in rads per second.

*w* = (71rev/min) (1min/60s) (2(pi)rad/1rev)

*w *= 7.43rad/s

a) The tangential acceleration *a* is given by a=r*a*. We want this radius in meters.

r = (7.5in) (2.54cm/1in) = 19.05cm = 0.1905m

a* = *r*a = *r*w*/t = (0.1905m) (7.43rad/s) / (5.2s) = 0.272 m/s^2

b) The tangential velocity v is given by v=r*w*

v = (0.1905m) (7.43rad/s) = 1.416m/s

c) First we need the angular velocity *w* after 1 second

*w = w*i + *a*t = 0 + (7.43rad/s / 5.2s) (1s) = 1.429rad/s

Radial acceleration *a*c = r*w*^2 = (0.1905m) (1.429rad/s)^2 = 0.389rad/s^2

Now total acceleration Atot is given by the distance formula with *a*c and *a* from earlier.

Atot = sqrt[(0.272)^2 + (0.389)^2] = 0.475m/s^2

d) The angle *x* that the total acceleration makes with *a*c is given by

*x* = tan^-1(*a/a*c) = tan^-1(0.272/0.389) = 34.96degrees

Proudly made in USA and serving the world! © Akari Online, 2020