An athlete whirls a 7.54 kg hammer tied to the end of a 1.5 m chain in a horizontal circle.
The hammer moves at the rate of 1.51 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?
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 velocity, and r is the radius of the circular path.
Given that the hammer moves at a rate of 1.51 revolutions per second and the length of the chain is 1.5 m, we can calculate the velocity using the formula:
v = 2πr/T
where T is the period or the time it takes for one revolution (T = 1/f). In this case, the frequency f is 1.51 rev/s. Therefore, the period T will be:
T = 1 / 1.51
Substituting this value into the velocity formula, we get:
v = 2π(1.5 m) / (1 / 1.51) = 9.525 m/s
Next, we can substitute the values of v and r into the centripetal acceleration formula:
ac = (9.525 m/s)^2 / 1.5 m = 60.252 m/s^2
So, the centripetal acceleration of the hammer is 60.252 m/s^2.
To find the tension in the chain, we can use the following equation:
Tension = (mass × centripetal acceleration) + (mass × gravitational acceleration)
Given that the mass of the hammer is 7.54 kg and the gravitational acceleration is approximately 9.8 m/s^2, we can substitute the values in:
Tension = (7.54 kg) × (60.252 m/s^2) + (7.54 kg) × (9.8 m/s^2)
Tension = 457.435 kg.m/s^2 + 73.892 kg.m/s^2
Simplifying further, we get:
Tension = 531.327 kg.m/s^2
So, the tension in the chain is approximately 531.327 kg.m/s^2.