An athlete whirls a 6.6 kg hammer tied to the end of a 1.2 m chain in a simple horizontal circle where you should ignore any vertical deviations. The hammer moves at the rate of 1.02 rev/s.
What is the centripetal acceleration of the hammer? Assume his arm length is included in the length given for the chain.
Answer in units of m/s2.
What is the tension in the chain?
change rev/s to radians/sec
centacceleration-masshammer *w^2*rad
For some reason I didn't realize that revolutions/sec was ANGULAR frequency. Thank You!
one revolution is 2 pi radians
so
one Rev/second = 2 pi radians/second
To find the centripetal acceleration of the hammer, we can use the formula:
ac = (v^2) / r
where:
ac is the centripetal acceleration,
v is the linear velocity of the object,
and r is the radius of the circular path.
In this case, the linear velocity isn't given directly, but we can calculate it using the given information. The hammer completes 1.02 revolutions per second, which means it travels 1.02 x 2π radians per second.
To get the linear velocity, we can multiply the angular velocity by the radius of the circular path. In this case, the radius is the length of the chain, which is given as 1.2 m.
v = 1.02 rev/s x 2π rad/rev x 1.2 m = 7.63 m/s
Now we can substitute the values into the centripetal acceleration formula:
ac = (7.63 m/s)^2 / 1.2 m = 48.44 m/s^2
Therefore, the centripetal acceleration of the hammer is 48.44 m/s^2.
To find the tension in the chain, we can use the formula:
T = (m x ac) + mg
where:
T is the tension in the chain,
m is the mass of the hammer,
ac is the centripetal acceleration,
and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The mass of the hammer is given as 6.6 kg. Substituting these values into the equation:
T = (6.6 kg x 48.44 m/s^2) + (6.6 kg x 9.8 m/s^2)
T = 317.904 kg m/s^2 + 64.68 kg m/s^2
T = 382.584 kg m/s^2
Therefore, the tension in the chain is 382.584 kg m/s^2, or simply 382.584 N.