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professor190 [17]
3 years ago
11

A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the

ride turns. They are on the same radial line. At one instant, the acceleration of the purse is . At that instant and in unit-vector notation, what is the acceleration of the wallet?
Physics
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

The acceleration of the wallet is 3\hat{i}+6\hat{j}

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse a=2\hat{i}+4.00\hat{j}

We need to calculate the acceleration of the wallet

Using formula of acceleration

a=r\omega^2

Both the purse and wallet have same angular velocity

\omega=\omega'

\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}

\dfrac{a}{r}=\dfrac{a'}{r'}

\dfrac{a'}{a}=\dfrac{r'}{r}

\dfrac{a'}{a}=\dfrac{3.45}{2.30}

\dfrac{a'}{a}=\dfrac{3}{2}

a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})

a'=3\hat{i}+6\hat{j}

Hence, The acceleration of the wallet is 3\hat{i}+6\hat{j}

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Answer:

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Explanation:

In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:

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You replace the values of m, r and v in the equation for Fc:

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