Respuesta :
Answer:
a. [tex]a=0.06m/s^2[/tex]
b. T1 = 4.62N
c. T2 = 4.97N
d. [tex]\alpha=1.20rad/s^2[/tex]
e. [tex]I = 0.0146 kg.m^2[/tex]
Explanation:
This question is incomplete. The original question was:
Block 1 has a mass m1 = 460 g, block 2 has mass m2 = 500 g, and the pulley, which is mounted on a horizontal axle with negligible friction,
has radius R = 5.00 cm. When released from rest, block 2 falls 0.750 m in 5.00 s without the cord slipping on the pulley.
(Do not treat the pulley as a uniform disk. Give your answers to three significant figures. Use these rounded values in subsequent calculations.)
a. What is the magnitude of the acceleration of the blocks?
b. What is the tension in the rope on block 1?
c. What is the tension in the rope on block 2?
d. What is the magnitude of the pulley's angular acceleration?
e. What is the rotational inertia?
We will use kinematics to calculate the acceleration of the blocks:
[tex]Y=Vo*t-1/2*a*t^2[/tex]
Where Y=-0.75m; Vo = 0m/s; t = 5.00s
Solving for a:
[tex]a=-0.06m/s^2[/tex] The magnitude of the acceleration is [tex]0.06m/s^2[/tex] and the acceleration of each block is
[tex]a1=0.06m/s^2[/tex]; [tex]a2=-0.06m/s^2[/tex]
By making a sum of forces on the block 1:
[tex]T1-m1*g=m1*a1[/tex] Solving for T1 and using [tex]g=10m/s^2[/tex]:
T1 = 4.62N
By making a sum of forces on the block 2:
[tex]T2-m2*g=m2*a2[/tex] Solving for T2 and using [tex]g=10m/s^2[/tex]:
T2 = 4.97N
Angular acceleration is given by:
[tex]\alpha=a/R=1.2rad/s^2[/tex]
By making a sum of torques on the pulley, we get the inertia:
[tex]T1*R-T2*R=-I*\alpha[/tex] Solving for I:
[tex]I=\frac{-R*(T1-T2)}{\alpha}[/tex]
[tex]I=0.0146kg.m^2[/tex]
a) [tex]\rm a_1 = 0.06\;m/sec^2\;\;and \;\; a_2 = - 0.06\;m/sec^2[/tex]
b) [tex]\rm T_1 = 4.62\;N[/tex]
c) [tex]\rm T_2 = 4.97\;N[/tex]
d) [tex]\rm \alpha = 1.2 \;rad/sec^2[/tex]
e) [tex]\rm I = 0.0146\;Kg.m^2[/tex]
Given :
Mass, [tex]\rm m_1 = 460\;g[/tex]
Mass, [tex]\rm m_2 = 500\;g[/tex]
Radius, R = 5 cm
When released from rest, block 2 falls 0.750 m in 5 sec without the cord slipping on the pulley.
Solution :
We know that,
[tex]\rm S = ut + \dfrac{1}{2}at^2[/tex]
Now put the values of u, S and t and solve for a we get,
[tex]\rm -0.75 = 0 +\dfrac{1}{2}a(5^2)[/tex]
[tex]\rm a = -0.06\;m/sec^2[/tex]
So, acceleration of each block is,
[tex]\rm a_1 = 0.06\;m/sec^2\;\;and \;\; a_2 = - 0.06\;m/sec^2[/tex]
Now for block 1,
[tex]\rm T_1-m_1g = m_1a_1[/tex] ----- (1)
Now put the values of [tex]\rm a_1,\;m_1\;and\;g[/tex] in equation (1),
[tex]\rm T_1 = 4.62\;N[/tex]
Now for block 2,
[tex]\rm T_2-m_2g = m_2a_2[/tex] ----- (2)
Now put the values of [tex]\rm a_2,\;m_2\;and\;g[/tex] in equation (2),
[tex]\rm T_2 = 4.97\;N[/tex]
Now for angular acceleration,
[tex]\rm \alpha = \dfrac{a}{R}[/tex]
[tex]\rm \alpha = 1.2 \;rad/sec^2[/tex]
Now sum of torques on the pulley is,
[tex]\rm T_1R-T_2R = -I\alpha[/tex]
[tex]\rm I = \dfrac{R(T_2-T_1)}{\alpha }[/tex]
[tex]\rm I = 0.0146\;Kg.m^2[/tex]
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https://brainly.com/question/6855614?referrer=searchResults
