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natasha13100
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Could someone please explain to me where I am going wrong? I've tried combining the relevant equations below in multiple ways. Also, I'm not sure if my answer is given in rad/s2 or °/s2.
A uniform solid disc of mass 1.2 kg and radius 0.24 m is free to rotate on a horizontal frictionless axle passing through the center of the disc. A long light string is wrapped around the disc; a block of mass 1.9 kg is suspended on the string as shown. The system is released from rest, and the block begins to descend. Find the magnitude of the angular acceleration α of the disc. Enter your answer in rad/s2 and use g = 9.8 m/s2.
force(F)=mass(m)*acceleration(a)
torque(t)=radius(R)*F*sin(θ) where θ=angle between R and F when the tails are placed together
Ʃt=moment of inertia(I)*angular acceleration(α)
I=mR2
tangential acceleration(atan)=Rα
I=1/2mR2/SUP] for discs of uniform density
I use m for the mass of the block and M for the mass of the disc. a is the tangential acceleration/acceleration of the block and α is the angular acceleration.
For the disc, there is a normal force exerted by the axle that is equal in magnitude and opposite in direction to the force of gravity. There is also string tension(T).
Two forces are acting on the mass: string tension and gravitational force.
The two equations relevant to α are Ʃt=I*α and a=R*α.
All of the torque is coming from the force T so Ʃt=t=RFsin(θ)=.24T. Additionally, I=1/2MR2=1/2(1.2)(.24)2=.03456. Therefore, .24T=.03456α and T=.144α
Tension also acts on the mass.
The net force(F)=T-the force due to gravity(G)=T-mg and F=ma so ma=T-mg and T=ma+mg=1.9a+1.9*9.8=1.9a+18.62
Therefore, .144α=1.9a+18.62. Next, I use a=Rα to make the equation .144α=1.9(.24)α+18.62 or .144α=.456α+18.62.
.312α=-18.62 so α=-4655/78≈-59.67948718
Homework Statement
A uniform solid disc of mass 1.2 kg and radius 0.24 m is free to rotate on a horizontal frictionless axle passing through the center of the disc. A long light string is wrapped around the disc; a block of mass 1.9 kg is suspended on the string as shown. The system is released from rest, and the block begins to descend. Find the magnitude of the angular acceleration α of the disc. Enter your answer in rad/s2 and use g = 9.8 m/s2.
Homework Equations
force(F)=mass(m)*acceleration(a)
torque(t)=radius(R)*F*sin(θ) where θ=angle between R and F when the tails are placed together
Ʃt=moment of inertia(I)*angular acceleration(α)
I=mR2
tangential acceleration(atan)=Rα
I=1/2mR2/SUP] for discs of uniform density
The Attempt at a Solution
I use m for the mass of the block and M for the mass of the disc. a is the tangential acceleration/acceleration of the block and α is the angular acceleration.
For the disc, there is a normal force exerted by the axle that is equal in magnitude and opposite in direction to the force of gravity. There is also string tension(T).
Two forces are acting on the mass: string tension and gravitational force.
The two equations relevant to α are Ʃt=I*α and a=R*α.
All of the torque is coming from the force T so Ʃt=t=RFsin(θ)=.24T. Additionally, I=1/2MR2=1/2(1.2)(.24)2=.03456. Therefore, .24T=.03456α and T=.144α
Tension also acts on the mass.
The net force(F)=T-the force due to gravity(G)=T-mg and F=ma so ma=T-mg and T=ma+mg=1.9a+1.9*9.8=1.9a+18.62
Therefore, .144α=1.9a+18.62. Next, I use a=Rα to make the equation .144α=1.9(.24)α+18.62 or .144α=.456α+18.62.
.312α=-18.62 so α=-4655/78≈-59.67948718