Question

Given: Consider the cart A with mass mA that rolls with respect to the ground (see the figure below). You may assume that the car is wide enough so that it cannot tip. A disk of radius R and mass mo is attached to the shaft on the cart at the center o of the disk (i.e., O is a pin joint). The disk is in contact with a horizontal surface at its top surface. A force F is applied to the cart when it is initially at rest. As the cart moves, the disk rolls without slipping. In your analysis, you may use mo = 20 kg, ma = 70 kg, R = 1 m, F = 100 N. no slip mo ma Find: The angular acceleration of the disk. Work on the white space provided. Once you are done, copy your answer in this box:

Answer 1

M FBD of Disc mo Due to no slip condition Ta=r - F mat 2 TN mag Force Balance in x-direction FBD of Cart AR ma= R, -f-0 Force Balance in Edirection F N+mg = R, - x-direction Force balance Torque Balance egn (about o) ma = F-F,-F, - R-0 T=fr = IX T-direction force balance fa 2 m, - к 1. mg+R, = N,+ N₂ - Torque Balance about Oci- If = moral RR = N,WE - NEW ©s @ - Rz 3 mord N, -M₂ = 2 R₂ r 38 O fit f Ema-k mra = { m Rx - ufen,+Nz)-(. may luCN4N2) = F -ma-Rx - 2m, 4 + N + 8 = N +N, E-ma-3 max = 1 more the (motMag N = (N,+N) - mg-mg-② 1 E-mara = 2 m rx tle (matm L2 Filom® 0 Ele(mot ma) g - x r (mA +2m) Is scanned with CamScanner

If you have any doubt in force balance of Cart check below.

FBD and Force Balance of the Cart:-

L. R = m g+N F TENA Y-axes Force balance N + N = m. g+N+mag X-ases Force balance Ry 4541, = F-ma Moment Balance R & R A N w -N, W zo ER = N, -N₂) w sov ZR w 2 - mg - mag W (NEN) - F- ma- Ry N,+N = IF-ma-R) ए: टरि CS Scanner with CaņScanner W

Where R1 and R2 are the reaction forces on the disc.

Also the height of the cart is considered to be very small when compared to the Radius of the Disc.