Question

a cyclist going round a circular with constant speed 10 km/hr complete 42 revolutions in 30 minutes. find its centripetal acceleration

Answer 1

Answer:

Its centripetal acceleration is 0.408 meter per second square, which equals to 5291 kilometer per hour square.

Explanation:

Refer to the pic...

Answer 2

The centripetal acceleration of the cyclist is $0.378 m/sec^{2}$.

Given,

Velocity of the cyclist=10 km/h

Number of revolutions made by the cyclist=42

Time taken for 42 revolutions=30 minutes.

To find,

the centripetal acceleration of the cyclist.

Solution:

• Frequency, f of an object is the number of revolutions made by the object in 1 second.
• The angular velocity, ω of an object is related with the frequency of that object as:
• ω=2πf.
• Centripetal acceleration arises due to the change in direction of the velocity of the object.
• It is directed towards the centre. It is given as:
• $a=\frac{v^{2} }{R}$ or $a=Rw^{2}$ or $a=vw$.
• v=speed of the object and R-radius of the path.

The frequency of the body will be:

Converting minutes into seconds:

$f=\frac{42}{30.60} \\f=0.023 Hertz.$

The angular velocity, ω of the body will be:

$w=2\pi f\\w=2.(3.14).(0.023)\\w=0.14 radians/sec.$

The centripetal acceleration of the cyclist will be:

Converting km/h into m/sec by multiplying it with 5/18.

$a=vw\\a=10X\frac{5}{18} X0.14\\a=10.(0.27).(0.14)\\a=0.378 m/sec^{2} .$

Hence, the centripetal acceleration is $0.378 m/sec^{2}$.

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