A body moves along a circular path with uniform angular speed of 0.6 rad s^-1 and at a constant speed of 3.0 ms^-1.

Calculate the acceleration of the body towards the centre of the circle.

Claire20

2 years ago

Firstly,find the displacement x using the formula V=wA
V = 3
w= 0.6
A= ?
3=0.6A
A = 5
Then to find the acceleration, use the formula a=-w²x ( where x is same as A= 5)
a= -0.6² × 5
a= 0.36 × 5
a= 1.8 ms-²

Overwatch886

2 years ago

To find the acceleration of the body towards the center of the circle, we can use the formula for centripetal acceleration:

\[a_c = \frac{v^2}{r}\]

Where:
- \(a_c\) is the centripetal acceleration,
- \(v\) is the speed of the body, and
- \(r\) is the radius of the circular path.

Given:
- \(v = 3.0 \, \text{m/s}\) (constant speed)
- \(\omega = 0.6 \, \text{rad/s}\) (angular speed)

We know that \(v = \omega \cdot r\), where \(r\) is the radius of the circular path. So, we can rearrange this equation to solve for \(r\):

\[r = \frac{v}{\omega}\]

Now, we can plug in the values to find \(r\):

\[r = \frac{3.0 \, \text{m/s}}{0.6 \, \text{rad/s}} = 5 \, \text{m}\]

Now, we can calculate the centripetal acceleration using the formula:

\[a_c = \frac{v^2}{r}\]

\[a_c = \frac{(3.0 \, \text{m/s})^2}{5 \, \text{m}}\]

\[a_c = \frac{9.0 \, \text{m}^2/\text{s}^2}{5 \, \text{m}}\]

\[a_c = 1.8 \, \text{m/s}^2\]

So, the acceleration of the body towards the center of the circle is \(1.8 \, \text{m/s}^2\).

thankgod320

6 years ago

a=wxspeed
a=3.0x0.6
a=1.8

johnteju

6 years ago

Explanation(s)
A = wv = 0.6 x 3.0 = 1.8m-2

chidera Malachy

9 years ago

Here is an explanation:

v=wa

where v=velocity

w=angular velocity

a=acceleration

therefore 3.0=0.6*a

a=3.0/0.6

a=1.8

REF: new school physics, page 167