Chapter 10, Problem 1Q

Figure 10-20 is a graph of the angular velocity versus time for a disk rotating like a merry-go-round, For a point on the disk rim, rank the instants a, b, c, and d according to the magnitude of the (a) tangential and (b) radial acceleration greatest first.

Figure 10-20 Question 1

To determine

To rank:

a) the tangential acceleration at given points a, b, c and d.

b) the radial acceleration at given points a, b, c and d.

Answer to Problem 1Q

Solution:

a) Ranking of tangential acceleration at given points is $c>a>b=d$.

b) Ranking of radial acceleration at given points is $b>a=c>d$.

Explanation of Solution

1) Concept:

From the slope of the graph we can rank the points according to the tangential acceleration. Then using the relation between radial acceleration and angular velocity we can rank the points according to the radial acceleration.

2) Formulae:

i) $a_{tangential}=r . \alpha$

ii) $a_{radial}={\omega }^2.r$

3) Given:

i. Graph of angular velocity vs time for a disk rotating like a merry-go-round.

ii. Points a, b, c and d are on the rim of the rotating disk.

4) Calculations:

a) As we know the formula for tangential acceleration as

$a_{tangential}=r . \alpha$

From the graph we can say that $\alpha$ is the slope of the $\omega$ vs time graph. The radius of the disk is same for all cases, so the value of tangential acceleration depends on angular acceleration. The slope is high at point c, so angular acceleration is maximum. After that point a has high slope and points b and d have zero slope.

 So the ranking is as,

$c>a>b=d$

b) The formula for radial acceleration is,

$a_{radial}={\omega }^2.r$

The radius of the disk is same, so the radial acceleration depends on value of $\omega$ which can be easily obtained from the graph. So ranking is,

$b>a=c>d$

Conclusion:

Observing the graph of angular velocity vs time we can find the radial and tangential acceleration of an object.