What part of Earth has a faster linear speed. *
1 answer:
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Answer:
The magnitude of the tangential velocity is
The magnitude of the resultant acceleration at that point is
Explanation:
From the question we are told that
The mass of the uniform disk is
The radius of the uniform disk is
The force applied on the disk is
Generally the angular speed i mathematically represented as
Where is the angular displacement given from the question as
is the angular acceleration which is mathematically represented as
The moment of inertial is mathematically represented as
Substituting values
Considering the equation for angular acceleration
Substituting values
Considering the equation for angular velocity
Substituting values
The tangential velocity of a given point on the rim is mathematically represented as
Substituting values
The radial acceleration at hat point is mathematically represented as
The tangential acceleration at that point is mathematically represented as
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The magnitude of resultant acceleration at that point is
Substituting values
Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
