Answer:
(1) 0.333 Hz
(2) 4 sec
(3) 2 sec, 0.5 Hz
Explanation:
(1) We have given time period of pendulum is 3 sec
So T = 3 sec
Frequency will be equal to
(2) Frequency of the pendulum is given f = 0.25 Hz
Time period is equal to
(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds
So time taken to complete 1 back and forth swings =
So time period T = 2 sec
Frequency will be equal to
Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
