Tangential Acceleration problem

[John O' Meara]

A light rigid rod 1.0m long has a small block of mass.05kg attached at one end. The other end is pivoted, and the rod rotates in a vertical circle. At a certain instant, the rod is 36.9 degrees above the horizontal, and the tangential speed of the block is 4m/s. (a) What are the horizontal and vertical components of the velocity of the block? (b) What is the moment of inertia of the block? (c) What is the radial acceleration of the block? (d) What is the tangential acceleration of the block? (e) What is the tension or compression in the rod?
Answers: (a) 3.2m/s and 2.4m/s.
(b) L=Iw => I=m*v*r/w=.05kg.m^2.
(c) a_radial=v^2/r=16 rad/s^2.
(d) a_tangential = r*(alpha), where alpha = angular acceleration. This is where my problem is.
(e) Fr=m*a_radial=.8N outwards. It is compressed by m*g*sin(36.9)=.294N, therefore T=.506N.
Where a_ is acceleration, Fr is radial force, w=angular velocity, L is angular momentum, mass=m, velocity = v. Many thanks.

 

[civil_dude]

And what is your question?

 

[John O' Meara]

My question is in part (d) What is the tangential acceleration of the block. If you know the answer please help. Thanks.

 

[Doc Al]

Consider Newton's 2nd law for rotational motion.

 

[PhanthomJay]

My question is in part (d) What is the tangential acceleration of the block. If you know the answer please help. Thanks.

The problem is that you have equated angular acceleration with centripetal acceleration. This is not correct; they are separate entities. Calculate the component of the weight in the direction of the tangent to the curve. This is the only force acting in that direction (the tension force acts perpendicular to it inward). Then use F=ma to solve for the tangential acceleration. Since F_t = mg sin theta = ma_t, then solve a_t = g sin theta.

 

[Doc Al]

The problem is that you have equated angular acceleration with centripetal acceleration.

Where did he do that?

 

[PhanthomJay]

Where did he do that?

where he said under (d)
a_t = (alpha)*r

 

[Doc Al]

where he said under (d)
a_t = (alpha)*r

That seems correct to me. (No mention of centripetal acceleration here.)

 

[John O' Meara]

Part (d), I know is a mistake as (alpha) means angular acceleration.

 

[Doc Al]

Part (d), I know is a mistake as (alpha) means angular acceleration.

Why is it a mistake?? (Tangential acceleration and angular acceleration are directly related.)

 

[PhanthomJay]

Why is it a mistake?? (Tangential acceleration and angular acceleration are directly related.)

Oh ya, correct, my mistake. It's just that you don't need to know alpha to find a_t in this problem. I think I made another mistake too; if the given angle is measured above the horizontal, the tangential component of the weight is mgcos theta, not mg sin theta. Sorry, bad day.

 

[John O' Meara]

I checked it out I was not wrong after all.Thanks for the help.