Respuesta :
Answer:
The magnitude of the tangential velocity is [tex]v= 0.868 m/s[/tex]
The magnitude of the resultant acceleration at that point is [tex]a = 4.057 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the uniform disk is [tex]m_d = 40.0kg[/tex]
The radius of the uniform disk is [tex]R_d = 0.200m[/tex]
The force applied on the disk is [tex]F_d = 30.0N[/tex]
Generally the angular speed i mathematically represented as
[tex]w = \sqrt{2 \alpha \theta}[/tex]
Where [tex]\theta[/tex] is the angular displacement given from the question as
[tex]\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}[/tex]
[tex]=1.257\ rad[/tex]
[tex]\alpha[/tex] is the angular acceleration which is mathematically represented as
[tex]\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}[/tex]
The moment of inertial is mathematically represented as
[tex]I = \frac{1}{2} m_dR^2_d[/tex]
Substituting values
[tex]I = 0.5 * 40 * 0.200^2[/tex]
[tex]= 0.8kg \cdot m^2[/tex]
Considering the equation for angular acceleration
[tex]\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}[/tex]
Substituting values
[tex]\alph[/tex][tex]\alpha = \frac{(30.0)(0.200)}{0.8}[/tex]
[tex]= 7.5 rad/s^2[/tex]
Considering the equation for angular velocity
[tex]w = \sqrt{2 \alpha \theta}[/tex]
Substituting values
[tex]w =\sqrt{2 * (7.5) * 1.257}[/tex]
[tex]= 4.34 \ rad/s[/tex]
The tangential velocity of a given point on the rim is mathematically represented as
[tex]v = R_d w[/tex]
Substituting values
[tex]= (0.200)(4.34)[/tex]
[tex]v= 0.868 m/s[/tex]
The radial acceleration at hat point is mathematically represented as
[tex]\alpha_r = \frac{v^2}{R}[/tex]
[tex]= \frac{0.868^2}{0.200^2}[/tex]
[tex]= 3.7699 \ m/s^2[/tex]
The tangential acceleration at that point is mathematically represented as
[tex]\alpha _t = R \alpha[/tex]
Substituting values
[tex]\alpha _t = (0.200) (7.5)[/tex]
[tex]= 1.5 m/s^2[/tex]
The magnitude of resultant acceleration at that point is
[tex]a = \sqrt{\alpha_r ^2+ \alpha_t^2 }[/tex]
Substituting values
[tex]a = \sqrt{(3.7699)^2 + (1.5)^2}[/tex]
[tex]a = 4.057 m/s^2[/tex]
(a) The tangential velocity of a point on the rim of the disk is
v = √10g(H−R)/7
(b) The magnitude of the resultant acceleration is given by
a = 10g(H−R)/(7R)
Rotational motion:
(a) From the law of conservation of energy:
PE(initial) + KE(initial) = PE(final) + KE(final) + RE
initialy Kinetic energy is 0
RE represents the rotational kinetic energy
mgH = mgR + ½mv² + ½Iω²
here I is the moment of inertia of the sphere
mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²
mgH = mgR + ½mv² + ⅕mv²
gH = gR + ⁷/₁₀ v²
v² = 10g(H−R)/7
v = [tex]\sqrt{10g(H-R)/7}[/tex] will be the tangential velocity of a point on the rim of the disk.
(b)The normal force exerted by the track on the sphere will be:
N = mv²/R
N = m (10g(H−R)/7) / R
N = 10mg(H−R)/(7R) = ma
So, acceleration is given by:
a = 10g(H−R)/(7R)
Learn more about conservation of energy:
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