bucket and pulley

logistic_guy

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A uniform solid cylindrical pulley of mass M=3 kg\displaystyle M = 3 \ \text{kg} and radius 20 cm\displaystyle 20 \ \text{cm} is mounted on a horizontal axle. The pulley is free to rotate, but a constant frictional torque of τf=0.6 Nm\displaystyle \tau_f = 0.6 \ \text{Nm} acts at the axle and opposes the pulley’s rotation. A light, inextensible, and non-stretching rope is tightly wrapped around the pulley. The rope has negligible mass and does not slip as it unwinds. A bucket of mass m=4 kg\displaystyle m = 4 \ \text{kg} is attached to the free end of the rope and hangs vertically. The system is released from rest at time t=0\displaystyle t = 0 and the bucket begins to fall.

Determine:

(a)\displaystyle \bold{(a)} The moment of inertia I\displaystyle I of the pulley.
(b)\displaystyle \bold{(b)} The linear acceleration a\displaystyle a of the bucket.
(c)\displaystyle \bold{(c)} The tension T\displaystyle T in the rope.
(d)\displaystyle \bold{(d)} The angular acceleration α\displaystyle \alpha of the pulley.
(e)\displaystyle \bold{(e)} The angular velocity ω\displaystyle \omega of the pulley at t=2 s\displaystyle t = 2 \ \text{s}.
(f)\displaystyle \bold{(f)} The linear velocity v\displaystyle v of the bucket at t=2 s\displaystyle t = 2 \ \text{s}.
 
A uniform solid cylindrical pulley of mass M=3 kg\displaystyle M = 3 \ \text{kg} and radius 20 cm\displaystyle 20 \ \text{cm} is mounted on a horizontal axle. The pulley is free to rotate, but a constant frictional torque of τf=0.6 Nm\displaystyle \tau_f = 0.6 \ \text{Nm} acts at the axle and opposes the pulley’s rotation. A light, inextensible, and non-stretching rope is tightly wrapped around the pulley. The rope has negligible mass and does not slip as it unwinds. A bucket of mass m=4 kg\displaystyle m = 4 \ \text{kg} is attached to the free end of the rope and hangs vertically. The system is released from rest at time t=0\displaystyle t = 0 and the bucket begins to fall.

Determine:

(a)\displaystyle \bold{(a)} The moment of inertia I\displaystyle I of the pulley.
(b)\displaystyle \bold{(b)} The linear acceleration a\displaystyle a of the bucket.
(c)\displaystyle \bold{(c)} The tension T\displaystyle T in the rope.
(d)\displaystyle \bold{(d)} The angular acceleration α\displaystyle \alpha of the pulley.
(e)\displaystyle \bold{(e)} The angular velocity ω\displaystyle \omega of the pulley at t=2 s\displaystyle t = 2 \ \text{s}.
(f)\displaystyle \bold{(f)} The linear velocity v\displaystyle v of the bucket at t=2 s\displaystyle t = 2 \ \text{s}.
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(a)\displaystyle \bold{(a)} The moment of inertia I\displaystyle I of the pulley.

We are now experts in moments of inertia and we have already derived the moment of inertia of a cylinder about its central axis.

The moment of inertia of the pulley is:

I=12MR2=12(3)(0.2)2=0.06 kgm2\displaystyle I = \frac{1}{2}MR^2 = \frac{1}{2}(3)(0.2)^2 = \textcolor{blue}{0.06 \ \text{kg} \cdot \text{m}^2}
 
(b)\displaystyle \bold{(b)} The linear acceleration a\displaystyle a of the bucket.

Gravity is pulling the bucket downward, so we can get our first equation as:

mgT=ma\displaystyle mg - T = ma

Two unknowns means that we cannot find the linear acceleration of the bucket right now. We then go to the other system. From the pulley, we can get:

RTτf=Iα=IaR\displaystyle RT - \tau_f = I\alpha = I\frac{a}{R}

Now we can take the tension in the rope T\displaystyle T from the first equation and substitute it in the second equation to get the linear acceleration a\displaystyle a.

T=mgma\displaystyle T = mg - ma

R(mgma)τf=IaR\displaystyle R(mg - ma) - \tau_f = I\frac{a}{R}

R2mgR2maRτf=Ia\displaystyle R^2mg - R^2ma - R\tau_f = Ia

Ia+R2ma=R2mgRτf\displaystyle Ia + R^2ma = R^2mg - R\tau_f

a=R2mgRτfI+R2m=(0.2)2(4)(9.8)(0.2)(0.6)0.06+(0.2)2(4)=6.58 m/s2\displaystyle a = \frac{R^2mg - R\tau_f}{I + R^2m} = \frac{(0.2)^2(4)(9.8) - (0.2)(0.6)}{0.06 + (0.2)^2(4)} = \textcolor{blue}{6.58 \ \text{m}/\text{s}^2}
 
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(c)\displaystyle \bold{(c)} The tension T\displaystyle T in the rope.
T=mgma=m(ga)=4(9.86.58)=12.88 N\displaystyle T = mg - ma = m(g - a) = 4(9.8 - 6.58) = \textcolor{blue}{12.88 \ \text{N}}
 
(d)\displaystyle \bold{(d)} The angular acceleration α\displaystyle \alpha of the pulley.
a=αR\displaystyle a = \alpha R

α=aR=6.580.2=32.9 rad/s2\displaystyle \alpha = \frac{a}{R} = \frac{6.58}{0.2} = \textcolor{blue}{32.9 \ \text{rad/}\text{s}^2}
 
(e)\displaystyle \bold{(e)} The angular velocity ω\displaystyle \omega of the pulley at t=2 s\displaystyle t = 2 \ \text{s}.
ω=ω0+αt\displaystyle \omega = \omega_0 + \alpha t

ω=0+(32.9)(2)=65.8 rad/s\displaystyle \omega = 0 + (32.9)(2) = \textcolor{blue}{65.8 \ \text{rad/s}}
 
(f)\displaystyle \bold{(f)} The linear velocity v\displaystyle v of the bucket at t=2 s\displaystyle t = 2 \ \text{s}.
v=v0+at\displaystyle v = v_0 + at

v=0+(6.58)(2)=13.16 m/s\displaystyle v = 0 + (6.58)(2) = \textcolor{blue}{13.16 \ \text{m/s}}
 
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