Answer:
Explanation:
Let T be the tension
For linear motion of hoop downwards
mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .
For rotational motion of hoop
Torque by tension
T x R , R is radius of hoop.
Angular acceleration be α,
Linear acceleration a = α R
So TR = I α
= I a / R
a = TR² / I
Putting this value in earlier relation
mg -T = m TR² / I
mg = T ( 1 + m R² / I )
T = mg / ( 1 + m R² / I )
mg / ( 1 + R² / k² )
Tension is less than mg or weight because denominator of the expression is more than 1.
Technically, this delivers a lot of energy into the Earth, but it’s
spread out over a large enough area that it doesn’t do much more than
leave footprints in a lot of gardens. A slight pulse of pressure spreads
through the North American continental crust and dissipates with little
effect. The sound of all those feet hitting the ground creates a loud,
drawn-out roar which lasts many seconds.
Answer:
spring deflection is x = (v2 / R + g) m / 4
Explanation:
We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration
Let's write the equations on the Y axis of this description
Fe - W = m 
Where Fe is elastic force, W the weight and the centripetal acceleration. The elastic force equation is
Fe = - k x
4 (k x) - mg = m v² / R
The four is because there are four springs, R is theradio of dip
We can calculate the deflection (x) of the springs
x = (m v2 / R + mg) / 4
x = (v2 / R + g) m / 4
