During a certain time interval, the angular position of a swinging door is descr

Question

During a certain time interval, the angular position of a swinging door is described by = 4.99 + 10.2t + 2.01t2, where is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door at the following times. (a) t = 0

(b) t = 3.01 s

a= rad/s2

2. Consider the system shown in the figure below with m1 = 23.0 kg, m2 = 13.6 kg, R = 0.190 m, and the mass of the pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1 is 4.80 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley.

(a)What is the magnitude of the acceleration of m1?
a = m/s2
(b) Calculate the time interval required for m1 to hit the floor.
t1 = s

(c) How would your answer change if the pulley were massless?

Please help, accurate answers, please

Explanation / Answer

Given

   equation is = 4.99 + 10.2t + 2.01t2 , theta is anglular displacement in rad, t is time in s

at time t=0 s

   = 4.99 + 10.2t + 2.01t2

t=0 , = 4.99 + 10.20*0 + 2.01*0^2
   = 4.99 rad

   w = d/dt = (d/dt)(4.99 + 10.2t + 2.01t2) = 0+10.2+2*2.01*t

t = 0 W = 10.2+2*2.01*0 = 10.2 rad/s

   = dW/dt = (d/dt)(10.2+2*2.01*t) = 0+4.02 = 4.02 rad/s2

b) at t = 3.01 s

   = 4.99 + 10.2t + 2.01t2
    = 4.99 + 10.20*3.01 + 2.01*3.01^2 rad
  
   = 53.902 rad

w   w = d/dt = (d/dt)(4.99 + 10.2t + 2.01t2) = 0+10.2+2*2.01*t

   at t = 3.01 s

   w = 10.2+2*2.01*3.01 rad/s = 22.30 rad/s
and

= dW/dt = (d/dt)(10.2+2*2.01*t) = 0+4.02 = 4.02 rad/s2

2.Given

   m1 = 23 kg, m2 = 13.6 kg, M = 5 kg, R = 0.190 m, h = 4.80m

as m1 is > than m2 so the acceleration of m1 is downward and m2 is upward
writing the force equation, with tensions in the string T1,T2 respective sides to m1,m2

for m1 is m1g - T1 = m1a

for m2 is T2 -m2g = m2a

  
adding

   T1-T2 = (m1-m2)g - (m1+m2)a ------------(A)

for the pulley rotaional motion the torque is

   (T1-T2)R = 0.5*MR^2*a/R (a= r*alpha, T = I*alpha)

   (T1-T2) = 0.5*Ma --------------(B)

equating (A)and (B)

   (m1-m2)g - (m1+m2)a = 0.5*Ma

   a = (m1-m2)g / (m1+m2+0.5*M)

   a = (23-13.6)9.8 /(23+13.6+0.5*5) m/s2

   a = 2.3560 m/s2

time taken to reach ground is

b)   s = ut+0.5 at^2

   4.80 = 0*t + 0.5*2.3560t^2

   t = 2.0186 s = 2.02 s

c) if pulley were massless then M = 0

   a = (m1-m2)g / (m1+m2+0.5*M)

   a = (23-13.6)9.8 /(23+13.6) m/s2

   a = 2.51693989 m/s2

time taken is
   s = ut+0.5 at^2

   4.80 = 0*t + 0.5*2.51693989t^2

   t =1.95299 s = 1.95299 s

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