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The tension in the rope when the gymnast climbs it at constant speed is 641.9 N.
Given:
Mass of gymnast, m = 65.5 kg
The speed 'v' of gymnast is constant
Solution:
Consider the free-body diagram of the system as shown below.
Balancing forces along the vertical axis we get:
ΣFy = 0
Thus, we get:
F = ma - (1)
where, m is mass of gymnast
a is acceleration of gymnast (a = 0m/s², as the speed is constant)
Also,
F = T - mg -(2)
where, T is tension in the rope
g is acceleration due to gravity
Equating (1) & (2), we get:
ma = T - mg
Re-arranging the equation, we get:
T = m(a+g)
Applying values in above equation we get:
T = (65.5 kg)(0 m/s²+9.8 m/s²)
T = 641.9 N
Therefore, the tension in the rope when the gymnast climbs it at constant speed is 641.9 N.
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Answer:
a_total = 2 √ (α² + w⁴) , a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 + ²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m