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CLASS ID#: (3) GIVEN (35 PTS): Two uniform disks A (m=10 kg) & B (my=6 kg) are fixed as one, pivoting about the frictionless
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Answer #1

Total mass moment of inertia of uniform disk;

I_o = \frac{1}{2}m_Ar_A^2 + \frac{1}{2}m_Br_B^2

\Rightarrow I_o = \frac{1}{2}(10)(0.200)^2 + \frac{1}{2}(6)(0.150)^2

\Rightarrow I_o =0.2675\;\;kg.m^2

For cylinder C;

\uparrow \sum F_y = m_ca_c

\Rightarrow T_c - 6(9.81)= m_c(\alpha \times r_A)

\Rightarrow T_c - 6(9.81)= 6(\alpha \times 0.200)

\Rightarrow T_c=58.86+ 1.2\alpha

For cylinder D;

\downarrow \sum F_y = m_Da_D

\Rightarrow 10(9.81) - T_D = m_D(\alpha \times r_B)

\Rightarrow 10(9.81) - T_D =10(\alpha \times 0.150)

\Rightarrow T_D = 98.1 - 1.5\alpha

For pulley;

(CW\;+)\;\;\;\;\sum M_o = I_o\alpha

\Rightarrow T_D r_B - T_C r_A = I_o\alpha

\Rightarrow ( 98.1 - 1.5\alpha)(0.150) - (58.86+ 1.2\alpha)(0.200)= 0.2675\alpha

\Rightarrow \alpha = 4.01775\;\;rad/s^2

(a)

Acceleration of C is;

a_c = \alpha \times r_A = 4.01775(0.20) = 0.80355\;\;m/s^2

Time required will be;

t = \frac{v-v_o}{a_C} = \frac{0.5-0}{0.80355}

\Rightarrow t =0.6222\;\;s

...(Answer)

(b)

Angular velocity will be;

\omega = \omega_o + \alpha t

\Rightarrow \omega = 0 + (4.01775) (0.6222)

\Rightarrow \omega =2.5\;\;rad/s

...(Answer)

(c)

Tension in C;

T_c=58.86+ 1.2\alpha

\Rightarrow T_c=58.86+ 1.2(4.01775)

\Rightarrow T_c=63.6813\;\;N

...(Answer)

Tension in D;

T_D = 98.1 - 1.5\alpha

\Rightarrow T_D = 98.1 - 1.5(4.01775)

\Rightarrow T_D = 92.0734\;\;N

...(Answer)

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